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Chapter 4 — Calculus

Class 11 · Mathematics

Overview

This unit introduces the basic ideas of calculus: limits, continuity and differentiation, and their early applications. You will learn what it means for a function to approach a value (limit), how to test whether a function behaves smoothly (continuity), and how to compute the rate at which a quantity changes (derivative). The unit explains the formal definition of derivative as a limit, standard differentiation rules, derivatives of trigonometric, exponential and logarithmic functions, and techniques like implicit differentiation. You will study higher order derivatives and learn how to use derivatives to find tangent and normal lines, test for increasing or decreasing behaviour of functions, and locate local maxima and minima. Simple problems on approximation and related rates show how calculus connects to physical situations. Mastering this unit builds a strong foundation for further calculus topics in class 12 and in science or engineering: derivatives are a language to describe change, and the tools here let you analyse motion, optimise quantities and prepare for integration.

Learning Objectives

  • Explain the concept of a limit and evaluate simple limits by algebraic methods
  • State and test continuity of a function at a point and on an interval
  • Define the derivative using first principles and compute derivatives of standard functions
  • Apply differentiation rules: sum, product, quotient and chain rules
  • Differentiate trigonometric, exponential and logarithmic functions and use implicit differentiation
  • Use first and second derivatives to determine tangents, normals, increasing/decreasing intervals and local extrema
  • Compute higher order derivatives and use differentials for linear approximation
  • Solve basic related rates and approximation problems using derivatives

Topics in this chapter

17 topics · tap a topic title to jump straight to it.

🔢1

Introduction to Calculus and Rates of Change

What calculus studies
Calculus is the branch of mathematics that formalises the idea of change and motion. At its core are two complementary themes: differentiation, which analyses instantaneous change, and integration, which accumulates quantities. In Class 11 we focus on the first theme. We study how functions change when their input changes, and how to describe that change using numbers called derivatives. These ideas appear in daily life: speed of a vehicle, growth rate of a population, and rate of chemical reactions all use the concept of rate.

From average to instantaneous rate
Begin with average rate: if a quantity y changes from y(x) to y(x+Δx) when the independent variable moves by Δx, the average rate is Δy/Δx. This gives a coarse idea of change over an interval. To get the instantaneous rate at x we make the interval very small and take the limit as Δx→0. This limiting process converts a ratio that depends on an interval into a single value attached to a point — the derivative. Understanding the limiting process is essential because many algebraic manipulations rely on it.

A simple motivating example
Consider distance s(t)=t^2 (in suitable units). Average speed from t to t+Δt is [(t+Δt)^2 − t^2]/Δt = 2t + Δt. As Δt becomes very small, the average speed approaches 2t. Thus the instantaneous speed at time t is 2t. The derivative transforms a formula for position into a formula for velocity.

Geometric and physical interpretations
Geometrically, the derivative at a point gives the slope of the tangent line to the curve y=f(x) at that point. This slope describes how steep the curve is and in which direction it moves. Physically, if f measures position, its derivative measures velocity; if f measures velocity, its derivative measures acceleration. The derivative therefore connects algebraic expressions to real-world change.

Role in problem solving
Calculus provides systematic tools: rules to find derivatives of many functions, tests to determine whether a function increases or decreases, and methods to locate maximum or minimum values under given conditions. These tools are used to approximate complicated expressions, to relate rates of change of quantities connected by constraints, and to create mathematical models. Developing skill with limits and derivatives in Class 11 prepares you for more advanced topics in Class 12 and for solving science and engineering problems.

Learning focus
In this chapter you will practice forming limits, applying the first principles definition of derivative, learning common differentiation rules, and using derivatives for tangents, optimisation and simple motion problems. Throughout, pay attention to domains, units (for example, trigonometric functions require radians in calculus), and the link between algebraic manipulation and geometric intuition. Regular practice with these ideas builds fluency and confidence.

📌 Examples
  • Distance s(t)=t^2; average speed from t to t+Δt is ( (t+Δt)^2−t^2 )/Δt = 2t+Δt ; instantaneous speed = 2t.
  • Population P(t)=100e^{0.03t}; rate of change at t is found by derivative which gives growth rate proportional to current size.
  • Temperature T(x) along a rod approximated linearly near x0 using differential: ΔT ≈ T'(x0)Δx.
🧮 Formulas
  1. Average rate of change = [f(x+Δx) − f(x)]/Δx
  2. \[Derivative (intuitive) f'(x)=lim_{Δx→0} [f(x+Δx) − f(x)]/Δx\]
📊 Visual ideas
Graph of y=f(x)=x^2 with secant line between x and x+Δx, and tangent line at x showing limiting position of secant as Δx→0
Sketch of a function with regions of increase and decrease indicated with arrows
🔢2

Limits: Definitions and Techniques

Formal idea of a limit
A limit studies the behaviour of a function as the input approaches a particular value. We write lim_{x→a} f(x)=L when f(x) can be made as close as desired to L by taking x sufficiently near a, excluding possibly x=a itself. This idea isolates the value the function 'approaches' rather than the function’s value at the point. Limits capture local behaviour and are the foundation of derivative definitions.

Useful intuition
Think of x moving along the horizontal axis towards a. The function values may oscillate or approach a stable value. If values get closer and closer to L from both left and right, the limit exists. If left-hand and right-hand approaches give different numbers, the two-sided limit does not exist. Limits also allow us to treat situations where the function itself is not defined at a point but behaves predictably nearby.

Algebraic methods for evaluation
Many limits are computed by algebraic manipulation. For rational expressions with both numerator and denominator approaching zero (0/0), factor polynomials and cancel common factors, then substitute. For expressions with square roots, multiply numerator and denominator by conjugates to eliminate radicals. For limits at infinity, divide by highest power of x in numerator and denominator to identify leading behaviour. Rationalisation, expansion, and simplification are standard tools.

One-sided limits and infinite limits
Left-hand limit lim_{x→a^-} and right-hand limit lim_{x→a^+} consider approach from one side only. A two-sided limit exists only when both one-sided limits match. If the function grows without bound as x approaches a, we write the limit as +∞ or −∞ and understand this as a vertical asymptote in a graph. Limits as x→∞ or x→−∞ describe end behaviour and horizontal asymptotes.

Standard special limits
Certain limits appear frequently and are essential to know: lim_{x→0} (sin x)/x = 1, lim_{x→0} (1+ x)^{1/x} = e, and limits comparing exponentials and polynomials at infinity (exponentials grow faster). These results are used in later proofs and derivative computations, especially for trigonometric and exponential functions.

Common pitfalls
Do not attempt to evaluate indeterminate forms by naive substitution without simplification. Indeterminate forms like 0/0, ∞/∞ require algebra or further limit techniques. Avoid mixing up limit existence with function value: a function can have a limit at a while being undefined or having a different value at a.

Practice approach
Start with substitution; if direct substitution gives a finite value, that is the limit. If you get an indeterminate form, try factorisation, conjugates or dividing by dominant powers. For oscillatory forms consider squeeze theorem where a function is trapped between two whose limits are known. With routine practice, choosing the right technique becomes quicker and more reliable.

📌 Examples
  • lim_{x→2} (x^2−4)/(x−2): factor numerator to get (x−2)(x+2)/(x−2)=x+2 → substitute x=2 gives 4.
  • lim_{x→0} (√(1+x)−1)/x: multiply by conjugate to get 1/(√(1+x)+1) → limit 1/2.
  • lim_{x→∞} (3x^2+5)/(2x^2−x)=3/2 by dividing numerator and denominator by x^2.
🧮 Formulas
  1. \[lim_{x→a} [f(x)+g(x)] = lim f + lim g (if limits exist)\]
  2. \[lim_{x→a} [f(x)g(x)] = (lim f)(lim g)\]
  3. \[lim_{x→a} f(x)/g(x) = (lim f)/(lim g) if lim g ≠ 0\]
📊 Visual ideas
Graph showing a function with a removable discontinuity: hole at x=a where limit exists but f(a) is different or undefined
Function with vertical asymptote at x=a where one-sided limits go to ±∞
🔢3

Continuity of Functions

Why continuity matters
Continuity describes functions without sudden jumps, holes or infinite spikes at a point. A continuous function at x=a behaves predictably: the value at a matches the value the nearby points approach. Many theorems in calculus — for example the Intermediate Value Theorem which guarantees a solution between two values — require continuity. Continuity is therefore both a practical property of physical models (no sudden jumps in temperature or position) and a technical requirement for deeper results.

Formal definition
Function f is continuous at a if three conditions hold simultaneously: f(a) is defined; lim_{x→a} f(x) exists; and lim_{x→a} f(x)=f(a). If any condition fails, the function is discontinuous at that point. Continuity on an interval means the function is continuous at every point of that interval. Continuity is always relative to the function’s domain: a function may be continuous on its domain but undefined outside it.

Types of discontinuities with examples
There are several kinds of discontinuities. A removable discontinuity occurs when the limit exists but the function is either undefined or has a different value at the point — a hole that can be 'filled'. For example (x^2−1)/(x−1) has a hole at x=1 but matches x+1 elsewhere. A jump discontinuity happens when left and right limits exist but are different; step functions are typical examples. An infinite discontinuity arises when values grow without bound near the point, producing a vertical asymptote, as in 1/(x−a).

Operations preserving continuity
Basic continuous functions include polynomials, sine and cosine, exponential and logarithmic functions on their natural domains. Important closure properties: sums, products and compositions of continuous functions are continuous (quotients are continuous where the denominator is non-zero). These facts allow construction of complex continuous functions from simpler building blocks, so verifying continuity often reduces to checking the domain and standard operations.

Testing for continuity
To test continuity at a point, evaluate the function at the point and compute the limit. If both match, continuity holds. For rational functions, check that denominator is non-zero; for piecewise functions check left and right limits. For removable discontinuities, sometimes redefining f(a) to equal the limit will make the function continuous, a useful trick in modelling or simplifying problems.

Continuity and differentiability
Differentiability implies continuity: if a function has a derivative at a point, it must be continuous there. However, the converse is false: continuous functions may fail to be differentiable at some points due to corners or cusps — for example f(x)=|x| is continuous everywhere but not differentiable at x=0. Understanding the distinction sharpens comprehension of what smoothness means in calculus.

Practical tips
In solving problems, identify domain first, then find points where algebraic expressions fail (denominator zero, negative inside even root, logarithm arguments ≤0) and check one-sided limits at boundaries. Use continuity to apply theorems, and to justify replacements of functions by simpler continuous approximations near points of interest.

📌 Examples
  • f(x)=(x^2−1)/(x−1) has a removable discontinuity at x=1; simplify to x+1 and define f(1)=2 to make it continuous.
  • Step function that equals 0 for x<0 and 1 for x≥0 has a jump discontinuity at x=0.
  • g(x)=sin x is continuous everywhere; h(x)=ln x is continuous for x>0 only.
🧮 Formulas
  1. \[f continuous at a ⇔ lim_{x→a} f(x)=f(a)\]
  2. If f and g are continuous at a, then f+g, fg and f/g (if g(a)≠0) are continuous at a
📊 Visual ideas
Graph with a removable hole at x=a and an assigned value elsewhere
Graph of step function showing jump at x=0
🔢4

Derivative: Definition by First Principles

Formal starting point
The derivative at a point is defined by a limit. For a function f, the derivative at x is f'(x)=lim_{h→0} [f(x+h)−f(x)]/h, provided the limit exists. This is the first-principles definition and is the rigorous statement of instantaneous rate of change. It measures how small changes in input produce small changes in output, scaled per unit input change.

Understanding the difference quotient
The expression [f(x+h)−f(x)]/h is the slope of the secant line passing through two points on the curve: (x,f(x)) and (x+h,f(x+h)). As h becomes very small, the secant approaches the tangent, and its slope approaches the derivative. Viewing the derivative as a limiting slope links algebraic manipulation to geometric intuition: the derivative is the slope of the tangent line at the point.

Step-by-step computation
To compute derivative from first principles: write the difference quotient, expand and simplify algebraically to cancel h where possible, and then take the limit as h→0. For polynomials and many elementary functions this process is straightforward and produces the familiar rules. For functions with absolute values, corners or cusps, left and right limits may differ, causing non-existence of derivative at that point.

Examples and special cases
For f(x)=x^2 the difference quotient becomes 2x+h and hence the derivative is 2x. For linear function f(x)=mx+c the difference quotient simplifies to m, so the derivative is m constant everywhere. For f(x)=|x| the left and right difference quotients at 0 are −1 and +1, respectively, so derivative does not exist at 0; the graph has a corner there. Such examples illustrate that differentiability is a stronger condition than continuity.

Local linearity and approximations
If f is differentiable at x0, then locally the function behaves like its tangent line: f(x) ≈ f(x0) + f'(x0)(x−x0) for x near x0. This linear approximation underpins many practical techniques: error estimation, approximating function values and starting iterative methods for solving equations. Differentiability therefore not only gives slope but also a local linear model of the function.

Implications and limitations
Differentiability implies continuity, but not vice versa. Points where derivative fails may indicate important geometric features such as cusps, corners or vertical tangents. In advanced topics, the first principles definition is used to prove differentiation rules and to justify properties like the Mean Value Theorem. Practically, computing derivatives by first principles is often slower but instructive for understanding why the rules work.

📌 Examples
  • f(x)=x^2: [ (x+h)^2−x^2 ]/h = 2x+h → limit h→0 gives f'(x)=2x.
  • f(x)=3x+5: difference quotient simplifies to 3 → derivative f'(x)=3 constant.
  • f(x)=|x| at x=0: left-hand derivative = −1, right-hand derivative = +1, so derivative does not exist at 0.
🧮 Formulas
  1. \[f'(x)=lim_{h→0} [f(x+h)−f(x)]/h\]
  2. Differentiable ⇒ Continuous
📊 Visual ideas
Curve y=x^2 with tangent at x=1 showing slope 2
Graph of y=|x| showing corner at origin where tangent slope jumps
🔢5

Basic Differentiation Rules

Overview of practical rules
While the first-principles definition gives the meaning of a derivative, in practice we rely on rules that allow quick computation for most functions. These rules are proved from the definition and include linearity (sum and constant multiple), the power rule, the product rule and the quotient rule. Learning and applying them correctly saves time and reduces algebraic work.

Linearity and power rule
The sum rule says the derivative of a sum is the sum of derivatives: d/dx[f(x)+g(x)]=f'(x)+g'(x). Constants pull out: d/dx[c f(x)]=c f'(x). The power rule is central: for integer n, d/dx[x^n]=n x^{n−1}. This rule also extends to rational exponents where the function is defined. Using these, differentiation of polynomials becomes mechanical: apply the power rule term by term and combine with linearity.

Product rule
If u(x) and v(x) are differentiable, the derivative of their product is d/dx[ u v ] = u' v + u v'. This is necessary because simply multiplying derivatives would ignore interaction terms; the product of two changing quantities changes both because each factor changes. The product rule is proved using first principles by expanding the difference quotient and regrouping terms.

Quotient rule
The derivative of a quotient u(x)/v(x) (with v≠0) is (u'v − u v')/v^2. It follows from the product rule by writing u/v = u·(1/v) and differentiating or directly from first principles. Remember to square the denominator and to apply the rule carefully to avoid sign errors. Domain restrictions matter: if v(x0)=0 the derivative via quotient formula is not valid at that point.

Combining rules and higher derivatives
Use the rules together: product rule with chain rule, or quotient with chain when functions are nested. Higher order derivatives are obtained by differentiating results again; for example the second derivative of a polynomial follows by reapplying the power rule. Notation varies: f', f'', f^{(n)}, dy/dx and d^2y/dx^2 are used; become comfortable with these forms as exercises and solutions will use them interchangeably.

Common pitfalls and tips
Be careful to identify u and v correctly in product and quotient situations; distribute signs correctly in quotient rule. For functions like (x^2+1)(sin x) or sin x / x^2, write u and v and apply the rules rather than expanding into approximations. For composite functions use chain rule (covered separately). Practice with mixed examples (polynomials times trigonometric or exponential) to build speed and accuracy.

📌 Examples
  • d/dx(5x^3 − 2x + 7) = 15x^2 − 2
  • If u=x^2 and v=sin x, d(uv)/dx = 2x sin x + x^2 cos x
  • d/dx( (x^2+1)/(x−1) ) = [2x(x−1)−(x^2+1)(1)]/(x−1)^2
🧮 Formulas
  1. d/dx [f+g] = f' + g'
  2. d/dx [cf] = c f'
  3. \[d/dx [x^n] = n x^{n-1}\]
  4. d/dx [uv] = u'v + uv'
  5. d/dx [u/v] = (u'v − uv')/v^2
📊 Visual ideas
Plot of polynomial and its derivative showing where derivative is zero at local extrema
Two curves u and v and their product curve with tangent slopes illustrated
📐6

Derivatives of Trigonometric Functions

Importance of trigonometric derivatives
Trigonometric functions model periodic phenomena such as waves, oscillations and rotations. Their derivatives appear frequently in physics and engineering. To differentiate trig functions we rely on special limits and identities. One must measure angles in radians for the derivative formulas to be correct without extra conversion factors — a crucial point often emphasised in exercises.

Derivative of sin and cos
The derivatives of sine and cosine are basic and lead many other formulas. Using trigonometric limits and identities, one shows d/dx(sin x)=cos x and d/dx(cos x)=−sin x. These relationships reflect the phase shift between sine and cosine: the slope of sin is cos, and the slope of cos is negative sine. Because of these properties, derivatives of sine and cosine are again trigonometric functions — a helpful closure property.

Derivatives of other trig functions
Other trig derivatives can be obtained from sin and cos using quotient rule. For tangent, tan x = sin x / cos x, so d/dx(tan x)=sec^2 x. For cot x = cos x / sin x, derivative is −csc^2 x. For sec x and csc x, differentiate using product with reciprocal or quotient: d/dx(sec x)=sec x tan x and d/dx(csc x)=−csc x cot x. Memorising these formulas is useful for quick problem solving.

Using chain rule with trig functions
Frequently trig functions appear with composite arguments like sin(3x) or cos(x^2+1). Use the chain rule: d/dx[sin(g(x))]=cos(g(x))·g'(x). Always multiply by derivative of the inner function. This combination of trig and chain rules is common in differentiation tasks and related rates problems.

Proving basic trig derivatives
Proofs typically use limits: lim_{h→0} (sin h)/h = 1 and lim_{h→0} (1−cos h)/h = 0. Using the definition f'(x)=lim_{h→0} [f(x+h)−f(x)]/h together with trigonometric addition formulas yields the derivative of sine and cosine. These proofs give confidence that trig derivatives follow logically from earlier material on limits and continuity.

Applications and caution about units
When differentiating trig functions in physics, remember that if the angle is given in degrees you must convert to radians by multiplying by π/180; otherwise an extra factor appears in the derivative. Trigonometric derivatives are applied in problems about oscillations, pendulums (small-angle approximations), and harmonic motion. Practice computing derivatives of mixed expressions to build intuition and prevent sign errors.

📌 Examples
  • d/dx(sin(2x)) = cos(2x)·2 = 2 cos(2x)
  • d/dx(tan x) = sec^2 x
  • d/dx( sin x · x^2 ) = cos x · x^2 + sin x · 2x
🧮 Formulas
  1. d/dx(sin x)=cos x
  2. d/dx(cos x)=−sin x
  3. d/dx(tan x)=sec^2 x
  4. d/dx(sec x)=sec x tan x
  5. d/dx(cot x)=−csc^2 x
  6. d/dx(csc x)=−csc x cot x
📊 Visual ideas
Graph of y=sin x and y=cos x showing phase shift and relation of slopes
Graph of y=tan x showing vertical asymptotes and where derivative sec^2 x is large
🔢7

Derivatives of Exponential and Logarithmic Functions

Why exponentials and logarithms matter
Exponential and logarithmic functions model growth and decay, compound interest, radioactive decay, and many continuous processes. Their derivatives reveal how these processes change instant by instant. The natural exponential base e is especially important because the derivative of e^x is itself, making calculus formulas simpler and widely used in modelling.

Derivative of e^x and a^x
For the natural exponential, d/dx(e^x)=e^x. For a general base a>0, a≠1, write a^x = e^{x ln a} and apply chain rule to obtain d/dx(a^x)=a^x ln a. Thus derivatives of exponentials are proportional to the original function, with the proportionality constant ln a for base a. This property leads to differential equations of the form dy/dt = ky having solutions y=Ce^{kt}.

Derivative of natural logarithm and general logs
The natural logarithm has derivative d/dx(ln x)=1/x for x>0. This follows from the inverse relationship between ln and e^x and implicit differentiation. For log base a, d/dx(log_a x)=1/(x ln a). Using chain rule, for composite arguments d/dx(ln(g(x))) = g'(x)/g(x). Logarithmic differentiation is a useful technique for differentiating functions where the variable appears both in base and exponent or in products of many factors.

Logarithmic differentiation technique
To differentiate y = [g(x)]^{h(x)}, take natural logarithm of both sides: ln y = h(x) ln g(x). Differentiate implicitly: y'/y = h'(x) ln g(x) + h(x) g'(x)/g(x). Then multiply by y to find y'. This technique converts complicated products and powers into manageable sums and products and is particularly helpful when both base and exponent depend on x.

Applications and practical notes
Exponential derivatives model continuous compound interest and population growth; logarithmic derivatives simplify relative change computations: d/dx ln f(x) = f'(x)/f(x), so the derivative of ln f expresses relative rate of change of f. In problems check domain x>0 for logs. Also when working with exponentials and trig or polynomial combinations, always apply chain rule carefully. The simple form of e^x derivatives often simplifies higher order derivatives: d^n/dx^n e^{ax} = a^n e^{ax}.

Examples to build intuition
Differentiate e^{3x}: get 3e^{3x}. Differentiate 2^x: get 2^x ln 2. Differentiate y=x^x using logarithmic differentiation: ln y = x ln x ⇒ y'/y = ln x + 1 so y' = x^x(ln x + 1). These procedures are essential tools in a student's calculus toolkit.

📌 Examples
  • d/dx(e^{3x}) = 3 e^{3x}
  • d/dx(2^x) = 2^x ln 2
  • Differentiate y = x^x by writing ln y = x ln x; then differentiate to get y'/y = ln x + 1 ⇒ y' = x^x(ln x + 1)
🧮 Formulas
  1. d/dx(e^x)=e^x
  2. d/dx(a^x)=a^x ln a
  3. d/dx(ln x)=1/x
  4. d/dx(log_a x)=1/(x ln a)
📊 Visual ideas
Graph of y=e^x showing rapid growth and its tangent at x=0 with slope 1
Graph of y=ln x showing slow growth and vertical asymptote at x=0
🔢8

Chain Rule and Differentiation of Composite Functions

Need for the chain rule
Many functions are composed: one function inside another. For example sin(3x^2) or (1+x^2)^5. The chain rule explains how to differentiate such composite functions by separating the outer and inner parts. It formalises how change propagates through nested functions: a small change in x affects the inner function, which in turn affects the outer function.

Chain rule statement and interpretation
If y = f(g(x)) with f and g differentiable, then dy/dx = f'(g(x))·g'(x). In words: differentiate the outer function at the inner function, then multiply by the derivative of the inner function. Repeated composition requires applying the rule multiple times.

How to apply it step-by-step
1) Identify inner function u=g(x) and outer function f(u). 2) Compute du/dx = g'(x). 3) Compute df/du evaluated at u=g(x). 4) Multiply: dy/dx = (df/du)(du/dx). For deeper nests like f(g(h(x))), set u=g(h(x)) and proceed, or apply the rule sequentially: f'(g(h(x)))·g'(h(x))·h'(x). Practise spotting inner functions quickly for fluency.

Examples and special cases
For y = (1+x^2)^5, let u=1+x^2; dy/du =5u^4 and du/dx =2x; thus dy/dx =5(1+x^2)^4·2x =10x(1+x^2)^4. For y = sin(3x^2), dy/dx = cos(3x^2)·6x. Composite exponentials like e^{sin x} differentiate to e^{sin x}·cos x. Each example shows the same pattern: outer derivative evaluated at inner times inner derivative.

Common pitfalls and checks
Do not forget to multiply by the inner derivative; this is the most common mistake. When inner function is a product or quotient, apply product or quotient rule inside the chain rule as needed. Keep track of parentheses and algebraic simplification after differentiation. For inverse functions remember that differentiating inverse expressions requires implicit differentiation as an alternative method when appropriate.

Use with other rules
Often you need to combine the chain rule with product, quotient or power rules. For example differentiate x^2 sin(x^3): treat it as product u·v where v=sin(x^3) requires chain rule. Start by applying product rule, then within it use chain rule for v'. The chain rule is thus a central technique that integrates with other differentiation rules to handle most practical functions.

Practice strategy
Work on identifying the outermost function first, then peel layers inward. Write u=g(x) explicitly when in doubt. With practice, applying the chain rule becomes second nature and makes differentiating complicated expressions straightforward and reliable.

📌 Examples
  • y = sin(3x^2): let u=3x^2, dy/dx = cos(3x^2)·6x = 6x cos(3x^2)
  • y = (1+x^2)^5: dy/dx = 5(1+x^2)^4·2x = 10x(1+x^2)^4
  • y = e^{ olinebreak sin x}: dy/dx = e^{sin x}·cos x
🧮 Formulas
  1. If y=f(g(x)) then dy/dx = f'(g(x))·g'(x)
  2. For nested f(g(h(x))) derivative = f'(g(h(x)))·g'(h(x))·h'(x)
📊 Visual ideas
Sketch of composite function y=sin(x^2) with tangent at a point illustrating steepness change due to inner function
Diagram showing outer function f(u) and inner u=g(x) with arrows indicating derivatives multiplied
🔢9

Implicit Differentiation

When functions are not explicit
Not all curves are given as y explicitly in terms of x. Equations like x^2 + y^2 = 25 or x^3 + xy + y^3 = 6 describe relationships between x and y implicitly. To find dy/dx (slope of tangent) we differentiate both sides of the equation with respect to x, treating y as a function of x. Each time we differentiate a term containing y, we multiply by dy/dx using the chain rule.

Step-by-step method
1) Differentiate each term with respect to x. For a term like y^n use d/dx[y^n] = n y^{n−1}·(dy/dx). For products with y, such as x·y, use the product rule and remember that d/dx(y) = dy/dx. 2) After differentiating, collect all terms containing dy/dx on one side, factor dy/dx and solve algebraically for dy/dx. 3) Simplify and, if required, evaluate at specific points using the original equation to find y coordinate(s).

Applications and interpretation
Implicit differentiation provides slopes of tangents for curves not given in y=f(x) form; it also handles inverse function differentiation and related rates problems where variables are functions of time. The resulting dy/dx gives the instantaneous rate at which y changes with x along the curve. If the denominator in the final expression is zero at a point, the tangent may be vertical (slope infinite) and should be interpreted using limits.

Examples and typical algebra
For circle x^2 + y^2 = 25 differentiate: 2x + 2y dy/dx = 0 ⇒ dy/dx = −x/y. This gives slope at any point on the circle except where y=0 (horizontal points), where slope is infinite if x≠0. For x^3 + xy + y^3 = 6, differentiate: 3x^2 + (1·y + x dy/dx) + 3y^2 dy/dx = 0 and solve for dy/dx to get an explicit formula. Algebraic care is needed when collecting terms and solving.

Implicit differentiation with trigonometric and exponential terms
When the implicit equation contains functions of y, apply chain rule to those terms. For example differentiate sin y with respect to x to get cos y · dy/dx. Similarly, d/dx(e^{y}) = e^{y}·dy/dx. Keep track of these factors and treat dy/dx as an unknown to be isolated algebraically after differentiation.

Practical tips and checks
Always verify that the point at which you evaluate the derivative lies on the original curve. If the original equation yields multiple y values for a given x, evaluate dy/dx separately for each appropriate y. For vertical tangents use one-sided derivatives or implicit differentiation to identify infinite slopes. Practise with a variety of implicit curves to become comfortable with product rule, chain rule and algebraic rearrangement in this context.

📌 Examples
  • Circle x^2+y^2=25: differentiate to get 2x+2y dy/dx=0 ⇒ dy/dx = −x/y.
  • For x^3+xy+y^3=6: differentiate: 3x^2 + (1·y + x dy/dx) + 3y^2 dy/dx = 0 ⇒ dy/dx = −(3x^2+y)/(x+3y^2).
  • If x^2 + sin y = 4, differentiate: 2x + cos y · dy/dx = 0 ⇒ dy/dx = −2x / cos y.
🧮 Formulas
  1. \[Differentiate y-terms with chain rule: d/dx[y^n] = n y^{n-1} (dy/dx)\]
  2. Solve algebraically to isolate dy/dx after differentiating both sides
📊 Visual ideas
Circle x^2+y^2=25 with tangent at a point showing slope −x/y
Curve with implicit expression and labelled tangent whose slope found by implicit differentiation
🔢10

Higher Order Derivatives and Notation

What higher order derivatives represent
Higher order derivatives are simply derivatives of derivatives. The first derivative f' measures rate of change, the second derivative f'' measures how that rate itself changes. In physics, if f is position, f' is velocity and f'' is acceleration. Third and higher derivatives also have interpretations in some contexts: for instance, the third derivative is called jerk and measures rapid changes in acceleration.

Notation and calculation
We write derivatives in several notations: f'(x), f''(x), f^{(n)}(x) for the nth derivative, or dy/dx, d^2y/dx^2, d^n y/dx^n. To compute higher order derivatives, differentiate the expression for the previous order using the usual rules. For polynomials this is straightforward and often produces factorial-like coefficients; for more complicated functions apply product, quotient and chain rules at each step as required.

Patterns and useful results
Certain functions show predictable patterns in their derivatives. Exponentials: d^n/dx^n e^{ax} = a^n e^{ax}. Trigonometric functions cycle: derivatives of sin x produce cos, −sin, −cos and back to sin in a four-step cycle. Polynomials eventually reduce to zero after a finite number of differentiations. Recognising such patterns simplifies computation and helps in solving differential equations later.

Concavity and inflection points
The second derivative indicates concavity of the graph: f''(x)>0 implies the graph is concave upward (cup-shaped) near x, and f''(x)<0 implies concave downward (cap-shaped). Points where f''(x)=0 and concavity changes sign are inflection points. Identifying concavity and inflection points is important in sketching graphs and understanding the geometry of curves.

Application examples and signs to watch
For f(x)=x^4 the sequence of derivatives f'=4x^3, f''=12x^2, f^{(3)}=24x, f^{(4)}=24, f^{(5)}=0 shows how orders reduce degree. For f(x)=e^{2x}, f^{(n)}(x)=2^n e^{2x}. Note sign patterns for trig functions: sin leads to alternating signs. When using higher derivatives in tests for maxima/minima, the sign of the second derivative at a stationary point helps classify it, but when f''=0 the test is inconclusive and higher order derivatives may be examined for classification in some contexts.

Practical computation tips
Keep algebra neat: each differentiation may increase algebraic complexity. Use simplification between steps to reduce error. For repetitive structures, write down the pattern (for example coefficients) and then generalise to nth derivative when possible. Understanding higher derivatives prepares you for series expansions and differential equations in later study.

📌 Examples
  • If f(x)=x^4, f'(x)=4x^3, f''(x)=12x^2, f^{(3)}(x)=24x, f^{(4)}(x)=24, f^{(5)}(x)=0.
  • For f(x)=e^{2x}, f^{(n)}(x)=2^n e^{2x}.
  • For f(x)=sin x: f'(x)=cos x, f''(x)=−sin x, f^{(3)}(x)=−cos x, f^{(4)}(x)=sin x.
🧮 Formulas
  1. \[f^{(n)}(x) = d/dx [ f^{(n−1)}(x) ]\]
  2. \[d^n/dx^n e^{ax} = a^n e^{ax}\]
📊 Visual ideas
Graph showing a function and its concavity change; mark an inflection point where f'' changes sign
Illustration of sin x and its first two derivatives to show cyclic behaviour
🔢11

Tangents and Normals to Curves

Definition and geometric idea
The tangent to a curve at a point is the straight line that best approximates the curve near that point. Its slope is the derivative at the point. The normal is the line perpendicular to the tangent at the same point. Together they describe local geometry and are useful in approximation, optimisation and solving geometric problems involving curves.

Equation of tangent line
For a differentiable curve y=f(x) and a point (x_0,f(x_0)), the tangent line has slope m=f'(x_0). Using point-slope form the tangent equation is y − f(x_0) = f'(x_0)(x − x_0). This line equals the first-order linear approximation to the curve at x_0. It is used to estimate function values near x_0 and to compute intersections with other lines.

Normal line and its slope
The normal line is perpendicular to the tangent. If the tangent slope is m, then the normal slope is −1/m provided m ≠ 0. The normal at (x_0,y_0) thus has equation y − y_0 = (−1/m)(x − x_0). If the tangent is vertical (slope infinite), the normal is horizontal and has slope 0. If the tangent is horizontal (m=0), the normal is vertical.

Tangents for implicit curves
When curves are given implicitly, find dy/dx by implicit differentiation and evaluate at the point to obtain slope. For example, circle x^2 + y^2 = r^2 gives dy/dx = −x/y, so the tangent slope at (x_0,y_0) is −x_0/y_0 and normal slope y_0/x_0. Ensure the given point lies on the curve before computing tangent and normal.

Applications of tangents and normals
Tangents are used in linear approximations, estimating small changes, and in geometry problems such as finding angles between curves using their tangents. Normals appear in optics (law of reflection), mechanics (normal forces), and curvature studies. In optimisation, tangent lines help visualise local behaviour and support arguments about maxima and minima.

Practical steps and checks
1) Verify the point lies on the curve. 2) Compute f'(x) or dy/dx and evaluate at the point to get slope. 3) Write line equations carefully using point-slope form. 4) Check special cases: vertical tangent if denominator in dy/dx is zero, horizontal tangent if derivative is zero. Practise with explicit and implicit curves to master both techniques.

📌 Examples
  • For y=x^3 at x=1, f'(1)=3 ⇒ tangent: y−1=3(x−1) ⇒ y=3x−2.
  • Circle x^2+y^2=25 at (3,4): dy/dx = −x/y = −3/4 ⇒ tangent y−4 = −3/4 (x−3). Normal slope = 4/3.
  • For y=√x at x=4, f'(x)=1/(2√x) ⇒ f'(4)=1/4 ⇒ tangent y−2 = 1/4 (x−4).
🧮 Formulas
  1. Tangent: y−y_0 = f'(x_0)(x−x_0)
  2. Normal slope = −1/f'(x_0) (if f'(x_0)≠0)
📊 Visual ideas
Curve y=x^2 with tangent and normal drawn at x=1 showing intersection and slopes
Circle with tangent at a point and normal passing through centre
🔢12

Increasing and Decreasing Functions; Stationary Points

Monotonicity and derivative sign
The first derivative f'(x) is the principal tool for determining whether a function increases or decreases. If f'(x)>0 for every x in an interval, then as x increases the function values increase there: the function is strictly increasing. If f'(x)<0 on an interval, the function is strictly decreasing. A point where f'(x)=0 or where f' does not exist is called a critical or stationary point; such points are important because they may be the location of local maxima or minima, or of other special behaviour like inflection or cusp.

First derivative test in detail
To use the first derivative test, find all critical points by solving f'(x)=0 and noting where f' is undefined. Then examine the sign of f' on intervals determined by these critical points. If f' changes from positive to negative when crossing a critical point x0 from left to right, f has a local maximum at x0. If f' changes from negative to positive, f has a local minimum. If there is no sign change, the critical point is not a local extremum; it might be a point of inflection or a plateau where the function briefly flattens but continues in the same direction.

Second derivative test and concavity
The second derivative f'' gives information about how the slope itself is changing. If f''(x0)>0 at a stationary point x0 (where f'(x0)=0), the graph is concave upward there and x0 is a local minimum. If f''(x0)<0, the graph is concave downward and x0 is a local maximum. If f''(x0)=0 the test is inconclusive; further analysis such as studying higher derivatives or the sign of f' around x0 is necessary. Concavity also helps identify inflection points where the curve changes its curvature from up to down or vice versa.

Inflection points and their identification
An inflection point is where concavity changes sign. Often this happens at points where f''(x)=0 or is undefined, but a change in sign must be checked on either side to confirm an inflection. Some functions have inflection points that are not stationary (f' nonzero there), so do not assume inflection requires f'=0. Marking inflection points is essential for accurate curve sketching and for understanding the geometry of the graph.

Global versus local extrema and endpoints
Local extrema are maxima or minima within a neighbourhood. Global (absolute) maxima or minima are the largest or smallest values over the whole domain or a specified closed interval. When the domain is a closed bounded interval, the global extrema occur at critical points or at the endpoints. Therefore, after finding and classifying internal critical points, always evaluate the function at the interval endpoints to determine absolute maxima or minima.

Practical method and common checks
1) Determine domain and compute f'(x). 2) Solve f'(x)=0 and note where f' is undefined. 3) Make a sign chart or test values in each interval to track f' sign. 4) Use first or second derivative tests to classify stationary points. 5) For closed intervals compare function values at critical points and endpoints for absolute extrema. Check units and physical meaning in applied problems (e.g., length or time must be positive). Clear diagrams and sign charts help present reasoning in exams.

Examples and pitfalls
Watch out for functions like f(x)=x^3 where f'(0)=0 but the point is neither max nor min; instead it is an inflection with horizontal tangent. For functions not differentiable at a point, use one-sided derivatives or limit arguments. Always interpret the sign of derivative in context: positive means increasing, negative means decreasing; a zero derivative alone does not guarantee an extremum.

📌 Examples
  • f(x)=x^3 has f'(x)=3x^2 which is zero at x=0, but f''(0)=0 and the point is neither max nor min (inflection point).
  • f(x)=x^2: f'(x)=2x, f'(0)=0 and f''(0)=2>0 ⇒ local (and global) minimum at x=0.
  • For f(x)=x^3−3x, f'(x)=3x^2−3=0 ⇒ x=±1. Test: f''(x)=6x so at x=1, f''(1)=6>0 (min), at x=−1 f''(−1)=−6<0 (max).
🧮 Formulas
  1. If f'(x)>0 on interval ⇒ f increasing there; if f'(x)<0 ⇒ f decreasing there
  2. First derivative test: sign change of f' determines local extremum
  3. Second derivative test: f'(x_0)=0 and f''(x_0)>0 ⇒ local min; f''(x_0)<0 ⇒ local max
📊 Visual ideas
Graph showing increasing region where slope positive and decreasing region where slope negative with critical points marked
Function with local max and min annotated showing f' sign changes
🔢13

Maxima and Minima: Applications and Optimisation

Finding extrema in practice
Optimisation is a primary use of differentiation: find maximum or minimum values of quantities under given constraints. The general steps are: model the situation with an appropriate function f(x) representing the quantity to optimise; find the domain from physical or problem constraints; compute derivative f'(x) and find critical points by solving f'(x)=0 or where f' is undefined; classify critical points using first or second derivative tests; and check endpoints for global extrema on closed intervals. This systematic method turns word problems into calculus tasks.

Constrained optimisation (single-variable)
Many real problems involve several variables linked by a constraint. In Class 11, reduce such problems to a single variable using the constraint. For instance, with a fixed perimeter rectangle, express the area in terms of one side. Substitute and differentiate the single-variable function to find optimum dimensions. Avoid premature elimination; ensure algebraic steps respect the domain (lengths must be positive).

Typical problem types
Common problems include: maximize area for given perimeter, minimize surface area for given volume, find dimensions of containers for efficient storage, and economic problems like maximising profit or minimising cost given relationships between price, demand and revenue. Translate words into equations carefully and check units and feasibility of solutions.

Use of derivative tests
Once critical points are found, use the second derivative test for quick classification when possible: f''(x0)>0 implies local minimum, f''(x0)<0 implies local maximum. If f''(x0)=0, use the first derivative test or check nearby values. For global maxima/minima on closed intervals, evaluate function at critical points and endpoints; the largest value is global maximum and smallest is global minimum.

Examples and algebraic steps
Example: rectangle with fixed perimeter P. Let sides x and y with 2(x+y)=P ⇒ y=P/2−x. Area A(x)=x(P/2−x)= (P/2)x − x^2. Differentiate: A'(x)=P/2 − 2x; solve A'=0 ⇒ x=P/4. So rectangle is a square for maximum area. Always substitute back to compute actual maximum value. Check boundary values (x→0 or x→P/2) to ensure interior critical point gives maximum.

Practical modelling cautions
Define variables clearly and draw diagrams when helpful. Consider domain restrictions (lengths positive). Look for symmetry in problems: many optima occur for equal dimensions (making shapes regular). Verify solutions in context, e.g., negative lengths or unrealistic sizes are invalid. Optimisation combines algebra, differentiation and reasoning; clear steps produce correct and well-explained answers in exams.

📌 Examples
  • Maximise area of rectangle with perimeter 20: let sides x and y with 2(x+y)=20 ⇒ y=10−x, area A=x(10−x)=10x−x^2. A'=10−2x=0 ⇒ x=5 ⇒ square 5×5 gives max area 25.
  • Minimise surface area of open-top box made by cutting squares of side x from corners of 30×20 sheet: express volume and surface area in x, take derivative and find optimal x in domain.
  • Maximise volume of cylinder inscribed in a cone: express cylinder radius and height in terms of one variable using similarity of triangles, then optimise.
🧮 Formulas
  1. Set derivative to zero: f'(x)=0 gives critical points
  2. Use constraint relations to reduce multi-variable problems to single variable before differentiating
📊 Visual ideas
Sketch of a quadratic concave down showing maximum at vertex where derivative zero
Diagram of rectangle with fixed perimeter and variable side x, showing how area varies
🔢14

Curve Sketching Using Derivatives

Purpose of curve sketching
Curve sketching combines algebraic work and derivative tests to draw accurate graphs of functions without plotting many points. Using derivatives we find where a function rises or falls, where it has turning points, how it curves (concavity), where it might have asymptotes, and where behaviour at infinity leads to horizontal or slant asymptotes. A systematic approach yields clear sketches suitable for exams and modelling.

Step-by-step strategy
1) Domain: determine where function is defined and note any exclusions. 2) Intercepts: calculate f(0) and roots f(x)=0. 3) Asymptotes: vertical asymptotes at points where denominator zero (and limit ±∞); horizontal or slant asymptotes via limits at ±∞. 4) First derivative: compute f', find critical points f'=0 or undefined, and determine sign of f' on each interval to find where function increases or decreases. 5) Second derivative: compute f'', find points where f''=0 or undefined and test sign to determine concavity and locate inflection points. 6) Combine information and sketch, ensuring correct shape and slopes at key points.

Using derivative information effectively
First derivative sign chart locates maxima and minima; second derivative sign chart shows concavity. When f'(x0)=0 and f''(x0)≠0, use the second derivative test for classification. If f''(x0)=0, examine sign changes in f' or higher derivatives. Mark tangent line slopes at key points to indicate steepness. For rational functions, consider multiplicity of roots: if a factor is squared, the graph touches and rebounds; if odd multiplicity, it crosses the axis.

Asymptotes and end behaviour
Vertical asymptotes occur at domain exclusions where the function goes to ±∞; check one-sided limits. Horizontal asymptotes are given by limits as x→±∞. If degree numerator is one more than denominator in rational functions, there may be slant (oblique) asymptotes found by polynomial division. Including asymptotes in a sketch shows global trends and helps place the curve accurately for large |x|.

Special features and singularities
Handle cusps, corners and vertical tangents by checking differentiability and limits. For piecewise functions, ensure continuity and derivative changes at boundary points. Inflection points where concavity changes should be marked and explained. Combining all features—intercepts, monotonicity, extrema, concavity and asymptotes—produces a reliable sketch that captures both local and global behaviour.

Practice approach
Apply the six-step method to polynomials, rationals and simple transcendental functions. Start with simpler cases and build to more complex ones like rational-trigonometric products. Good sketches include labelled critical points, asymptotes drawn as dashed lines, and arrows indicating behaviour to infinity. Clear labelling and explanation of steps earn full credit in exams.

📌 Examples
  • Sketch y=(x^2−1)/(x−1): simplify to x+1 except at x=1 (hole). Domain x≠1, line y=x+1 with a hole at (1,2).
  • Sketch y=x^3−3x: find f' and f'', locate extrema at x=±1 and inflection at x=0 and draw correct cubic shape.
  • Sketch y=1/(x−2): vertical asymptote at x=2, horizontal asymptote y=0, opposite signs on two sides.
🧮 Formulas
  1. \[Asymptotes: vertical where denominator→0\]
    \[horizontal via lim_{x→±∞} f(x)\]
  2. Use f' and f'' signs to determine monotonicity and concavity
📊 Visual ideas
Complete sketch of a cubic y=x^3−3x with marked stationary points and inflection
Rational function y=1/(x−2) showing asymptotes and curve branches
🔢15

Differentials and Linear Approximation

Idea of differential
When x changes by a small amount dx, the corresponding change in y=f(x) is Δy=f(x+dx)−f(x). The differential dy is defined as dy=f'(x) dx and serves as a linear approximation to Δy when dx is small. The notion of differential captures the main linear part of the change and ignores higher order small terms. This is the mathematical expression of approximating a curve by its tangent line locally.

Linear approximation formula
The linear approximation or tangent line approximation states: f(x+Δx) ≈ f(x) + f'(x) Δx for small Δx. This is useful for estimating function values near known points without computing exact values. The formula gives the first-order Taylor approximation and underlies many numerical and physical estimation techniques.

Practical applications: error estimation
If a measurement of x has uncertainty dx, the approximate absolute error in y is dy ≈ f'(x) dx. The relative error is dy/f(x). For example, if radius r of a circle is measured with small error, approximate change in area A=πr^2 is dA ≈ 2πr dr. Engineers use differentials to propagate measurement errors and to estimate tolerances in manufactured parts.

Examples of numerical approximation
To estimate √(9.1), use f(x)=√x at x0=9 where f'(x)=1/(2√x)=1/6. With Δx=0.1, estimate f(9.1) ≈ 3 + (1/6)(0.1) ≈ 3.0167. Similarly sin(0.05) ≈ 0.05 because derivative of sin at 0 is 1. For small angles, sin θ ≈ θ and tan θ ≈ θ in radians — a common approximation in physics and engineering.

Limitations and when to use
Linear approximation is valid when Δx is sufficiently small and higher order terms are negligible. For larger changes, second-order and higher terms become relevant and a quadratic or full Taylor series gives better accuracy. Always estimate the size of neglected terms if high precision is needed. For many practical tasks with small tolerances the linear approximation is quick and sufficiently accurate.

Connecting differentials to derivatives
Understanding dy=f'(x) dx clarifies notation and emphasises that dy and dx are related quantities: dy depends linearly on dx with slope f'(x). This viewpoint is helpful for integration and differential equations later. In Class 11, focus on computing dy given dx and f'(x), and on using linear approximation to estimate function values and propagate small errors.

📌 Examples
  • Approximate √(9.1): f(x)=√x at x=9, f'(x)=1/(2√x)=1/6 ⇒ Δx=0.1 ⇒ Δy≈(1/6)·0.1≈0.0167 ⇒ estimate 3.0167.
  • Estimate sin(0.05) ≈ 0.05 using f(x)=sin x and f'(0)=1.
  • If a cylinder radius r measured as 5±0.01 cm, approximate error in area A=πr^2: dA ≈ 2πr dr = 2π·5·0.01 ≈ 0.314 cm^2.
🧮 Formulas
  1. dy = f'(x) dx
  2. f(x+Δx) ≈ f(x) + f'(x) Δx
📊 Visual ideas
Function y=f(x) with tangent line at x0 showing how tangent approximates curve near x0
Illustration of small Δx and corresponding Δy and linearised dy on graph
🏃16

Related Rates and Simple Motion Problems

Concept and approach
Related rates problems involve several quantities that change with time and are related by an equation. The goal is to find the rate of change of one quantity given the rates of change of others. The method uses implicit differentiation with respect to time t: differentiate the relation connecting variables, then substitute known values and solve for the desired derivative. This ties together algebra, geometry and differentiation.

Systematic steps
1) Introduce symbols for quantities and label which vary with time, e.g., x(t), y(t). 2) Form an equation relating the variables (geometry, physics or other). 3) Differentiate both sides with respect to t, applying chain rule: d/dt[y]=dy/dt, and when differentiating y(x(t)) multiply by dy/dt. 4) Substitute known variable values and known rates to solve for the unknown rate. Keep units consistent and pay attention to signs (positive means increasing, negative decreasing).

Classic examples
Ladder sliding: ladder length L fixed; base distance x and height y satisfy x^2 + y^2 = L^2. Differentiate: 2x dx/dt + 2y dy/dt = 0 ⇒ dy/dt = −(x/y) dx/dt. This relates speed of base sliding out to speed of top sliding down. Another type: expanding circular ripple where area A=πr^2 and dr/dt is given; then dA/dt = 2πr dr/dt. Conical tank filling uses similarity to relate radius and height then differentiates volume formula V=(1/3)πr^2 h to find dh/dt from dV/dt.

Checking signs and reasonableness
Interpret sign of result carefully: negative means decreasing quantity. Always compute relevant variable values before substituting rates. Check limiting cases: for instance when y→0 in ladder problem, dy/dt magnitude may become large, indicating rapid motion near the wall. Physically impossible results (like negative lengths) indicate modelling or algebraic errors.

Complexities and common errors
For related rates involving more variables, eliminate extra variables via constraints before differentiating if convenient. Remember to differentiate all terms with respect to t, and never drop dy/dt when differentiating y-terms. Units provide a useful consistency check: if result has wrong dimensions, there's an error. Practice problems help in recognising common templates and equations to use.

Applications and exam focus
Related rates are widely used in motion problems, growth/decay contexts and geometry-based rates. In exams, show clear steps: define variables, write relation, differentiate, substitute values, and state final answer with units. This clarity demonstrates method even when arithmetic is involved.

📌 Examples
  • Ladder 5 m long with top falling; if foot moves away at 1 m/s when foot is 3 m from wall, find rate of change of height: x^2+y^2=25 ⇒ 2x dx/dt + 2y dy/dt = 0 ⇒ dy/dt = −(x/y) dx/dt = −(3/4)·1 = −3/4 m/s.
  • A circular ripple radius r increases; area A=πr^2. If dr/dt = 0.2 cm/s at some time when r=10 cm, then dA/dt = 2πr dr/dt = 2π·10·0.2 = 4π cm^2/s.
  • Water fills a conical tank: V=(1/3)πr^2 h; relate r and h using geometry and differentiate to find dh/dt given dV/dt.
🧮 Formulas
  1. Differentiate relation with respect to t: if F(x,y,...)=0 then dF/dt = F_x dx/dt + F_y dy/dt + ... = 0 (by chain rule)
  2. Example: x^2 + y^2 = const ⇒ x dx/dt + y dy/dt = 0
📊 Visual ideas
Ladder leaning against wall diagram with labelled x (base distance), y (height) and velocities dx/dt, dy/dt
Conical tank cross-section with similar triangles relating radius and height
🔢17

Indeterminate Forms and an Introduction to L'Hôpital's Rule

Indeterminate forms and the need for extra work
When evaluating limits by direct substitution you sometimes get expressions like 0/0 or ∞/∞. These are called indeterminate forms because they do not determine the limit without further analysis. For 0/0 the numerator and denominator both vanish near the point, so their leading behaviours must be compared. Algebraic simplification, factoring, conjugation or series expansion can resolve many such limits. L'Hôpital's Rule provides a powerful technique for certain indeterminate ratios.

Statement of L'Hôpital's Rule (basic form)
Suppose functions f and g are differentiable near a (except possibly at a) and lim_{x→a} f(x) = lim_{x→a} g(x) = 0 (or both ±∞), and g'(x) ≠ 0 near a. If lim_{x→a} f'(x)/g'(x) exists (finite or infinite), then lim_{x→a} f(x)/g(x) = lim_{x→a} f'(x)/g'(x). The rule can be applied repeatedly if the resulting limit remains indeterminate. Use the rule only when the original limit is an indeterminate form of type 0/0 or ∞/∞.

When and how to apply
First attempt algebraic simplification. If that fails and the indeterminate form is 0/0 or ∞/∞, differentiate numerator and denominator separately and take the limit of their ratio. Ensure derivatives satisfy the rule’s conditions; do not apply L'Hôpital to forms like 0·∞ or 1^∞ without converting them into quotient form. Sometimes transforming the expression (e.g., reciprocal or taking logarithm) makes L'Hôpital applicable.

Examples and practice patterns
A common limit is lim_{x→0} (sin x)/x = 1. L'Hôpital gives cos 0 / 1 = 1 if one treats sin x / x as 0/0 and differentiates. For lim_{x→0} (1−cos x)/x^2, L'Hôpital applied twice yields 1/2. For lim_{x→∞} ln x / x, differentiating gives (1/x)/1 = 1/x → 0, so the original limit is 0, showing logarithm grows much slower than linear functions.

Limitations and cautions
L'Hôpital's Rule is not universal: it requires differentiability and the correct indeterminate form. It cannot be used if denominator derivative equals zero near the point or if the transformed limit does not exist. Also avoid unnecessary use when algebraic simplification is simpler. In exams justify application by verifying hypotheses: show original limit is 0/0 or ∞/∞ and that derivatives exist in a neighbourhood.

Practice advice
Work through several examples of 0/0 and ∞/∞ limits, both algebraic and transcendental, and practise converting other indeterminate forms into quotient form. Over time you will learn to pick the most efficient method—algebraic simplification where possible and L'Hôpital's Rule when appropriate. This builds a reliable toolkit for evaluating challenging limits in calculus problems.

📌 Examples
  • lim_{x→0} (sin x)/x = 1 is a standard limit; sometimes used with L'Hôpital as lim sin x / x → cos x / 1 = 1.
  • lim_{x→0} (1−cos x)/x^2 = 1/2: using L'Hôpital twice yields limit 1/2.
  • lim_{x→∞} (ln x)/x = 0 by applying L'Hôpital: (1/x)/(1)=1/x→0.
🧮 Formulas
  1. L'Hôpital's Rule: if lim f = lim g = 0 or ±∞ then lim f/g = lim f'/g' (when latter exists)
  2. Indeterminate forms include 0/0 and ∞/∞
📊 Visual ideas
Graph illustrating function ratios approaching 0/0 with secant behaviours where derivative ratio gives limit
Plot of ln x / x tending to 0 as x increases

Key Concepts

Limit
The value that a function approaches as the input approaches a point.
Continuity
A function is continuous at a point if its limit at that point equals its value there.
Derivative
The instantaneous rate of change of a function, defined as a limit of difference quotients.
Differentiable
A function is differentiable at a point if its derivative exists there.
Power Rule
A rule stating d/dx(x^n)=n x^{n-1} for suitable n.
Product Rule
The derivative of a product uv is u'v+uv'.
Quotient Rule
The derivative of u/v is (u'v−uv')/v^2 when v≠0.
Chain Rule
Derivative of f(g(x)) is f'(g(x))·g'(x).
Implicit Differentiation
Differentiating equations involving x and y by treating y as a function of x.
Higher Order Derivative
Derivatives obtained by differentiating a derivative, e.g., f'' is the second derivative.
Tangent
Line that touches a curve at a point with same instantaneous slope.
Normal
Line perpendicular to the tangent at the same point on the curve.
Inflection Point
Point where concavity changes, typically where second derivative changes sign.
Linear Approximation
Using tangent line: f(x+Δx) ≈ f(x) + f'(x)Δx for small Δx.
Related Rates
Problems where related quantities change with time and derivatives relate their rates.
L'Hôpital's Rule
A method to evaluate limits of indeterminate form 0/0 or ∞/∞ by differentiating numerator and denominator.

Practice Questions

  1. Find the derivative of y=x^3−5x+2 at x=2. / x=2 पर y=x^3−5x+2 का अवकलन ज्ञात कीजिए।
    Show answer

    Differentiate: y'=3x^2−5. At x=2, y'=3·4−5=12−5=7. / अवकलन करें: y'=3x^2−5. x=2 पर y'=7।

  2. Evaluate lim_{x→0} (sin x)/x. / lim_{x→0} (sin x)/x का मान ज्ञात कीजिए।
    Show answer

    This standard limit equals 1. / यह मान 1 के बराबर है।

  3. Find the equation of the tangent to y=√x at x=4. / y=√x पर x=4 पर स्पर्शरेखा का समीकरण बताइए।
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    y'=1/(2√x). At x=4, slope m=1/4. Point is (4,2). Tangent: y−2=(1/4)(x−4) ⇒ y=(1/4)x+1. / y'=1/(2√x). x=4 पर ढलान 1/4, बिंदु (4,2). स्पर्शरेखा: y−2=(1/4)(x−4) ⇒ y=(1/4)x+1।

  4. If x^2+y^2=25, find dy/dx at (3,4). / यदि x^2+y^2=25 है, तो (3,4) पर dy/dx ज्ञात कीजिए।
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    Differentiate implicitly: 2x+2y dy/dx=0 ⇒ dy/dx=−x/y. At (3,4), dy/dx=−3/4. / आंशिक अवकलन करने पर 2x+2y dy/dx=0 ⇒ dy/dx=−x/y. (3,4) पर dy/dx=−3/4।

  5. Find intervals where f(x)=x^3−6x^2+9x is increasing or decreasing. / f(x)=x^3−6x^2+9x किस अंतराल पर बढ़ती या घटती है? बताइए।
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    Compute f'=3x^2−12x+9 = 3(x^2−4x+3)=3(x−1)(x−3). Critical points at x=1 and 3. Sign chart: for x<1 f'>0 (increasing); 1<x<3 f'<0 (decreasing); x>3 f'>0 (increasing). / f'=3(x−1)(x−3). नोड x=1 और x=3। x<1 पर f'>0 तो बढ़ती, 1<x<3 पर f'<0 तो घटती, x>3 पर वापस बढ़ती।

  6. A ladder 10 m long leans against a wall. If bottom slides away at 1 m/s when bottom is 6 m from wall, how fast is top moving downward? / 10 मीटर लंबी सीढ़ी दीवार से लगी है। जब तल 6 मीटर दूर है और वह 1 मीटर/सेकंड की गति से दूर जा रही है, तो शीर्ष कितनी तीव्रता से नीचे जा रहा है? बताइए।
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    Let x be base distance, y height. x^2+y^2=100. Differentiate: 2x dx/dt + 2y dy/dt =0 ⇒ dy/dt = −(x/y) dx/dt. Here x=6 so y=√(100−36)=8. dx/dt=1 ⇒ dy/dt = −(6/8)·1 = −3/4 m/s (downward). / x^2+y^2=100 से 2x dx/dt + 2y dy/dt=0 ⇒ dy/dt=−(x/y) dx/dt. x=6, y=8, dx/dt=1 ⇒ dy/dt=−3/4 m/s।

  7. Use first principles to show derivative of f(x)=x^2 is 2x. / प्रथम सिद्धांत से दिखाइए कि f(x)=x^2 का अवकलन 2x है।
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    Difference quotient: [ (x+h)^2−x^2 ]/h = [x^2+2xh+h^2−x^2]/h = 2x + h. Taking limit as h→0 gives 2x. / (x+h)^2−x^2)/h = 2x+h; h→0 पर मान 2x मिलता है।

  8. Evaluate lim_{x→∞} (3x^2+5)/(2x^2−x). / lim_{x→∞} (3x^2+5)/(2x^2−x) का मान खोजिए।
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    Divide numerator and denominator by x^2: lim = (3 + 5/x^2)/(2 − 1/x) → 3/2 as x→∞. / x^2 से भाग देने पर सीमा (3+0)/(2−0)=3/2।

  9. Find dy/dx if y = x^x. / यदि y = x^x हो तो dy/dx ज्ञात कीजिए।
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    Take ln: ln y = x ln x. Differentiate: y'/y = ln x + 1 ⇒ y' = x^x(ln x + 1). / ln लेकर ln y = x ln x, प्रतिष्ठापन करने पर y'/y = ln x + 1 ⇒ y' = x^x(ln x + 1)。

  10. Determine whether f(x)=|x| is differentiable at x=0. / क्या f(x)=|x| x=0 पर अवकलनीय है? बताइए।
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    Left derivative at 0 is −1, right derivative is +1, they differ so derivative does not exist at 0. Thus not differentiable at 0. / बाएँ अवकलज −1 और दाएँ अवकलज +1 हैं; भिन्न होने पर अवकलन मौजूद नहीं है।

  11. Use linear approximation to estimate √(49.5). / रैखिक निकटता से √(49.5) का मान अनुमान लगाइए।
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    Let f(x)=√x at x0=49. f'(x)=1/(2√x) ⇒ f'(49)=1/14. Δx=−0.5. Approx: f(49.5)≈7 + (1/14)(0.5) =7 + 0.035714... ≈7.0357. / f'(49)=1/14, Δx=0.5 ⇒ 7 + (1/14)·0.5 ≈7.0357।

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