Differentiation

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Differentiation — Quick Facts Derivative as slope: dy/dx = lim(h→0) [f(x+h) - f(x)]/h Standard derivatives: d/dx(xⁿ) = nxⁿ⁻¹; d/dx(sin x) = cos x; d/dx(cos x) = -sin x; d/dx(eˣ) = eˣ; d/dx(ln x) = 1/x Chain rule: d/dx[f(g(x))] = f’(g(x)) · g’(x) Product rule: d/dx(uv) = u’v + uv’; Quotient rule: d/dx(u/v) = (u’v - uv’)/v² ⚡ Exam tip: In JEE, most differentiation questions test chain rule and implicit differentiation — master these

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Differentiation — Study Guide

Derivative Definition and Interpretation

The derivative of f(x) at point x is defined as: $$f’(x) = lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \frac{dy}{dx}$$

Geometrically: dy/dx represents the slope of tangent line to curve y = f(x) at point (x, f(x)). If slope = 0, point is stationary (maximum or minimum). If slope is undefined (vertical tangent or cusp), the derivative does not exist at that point.

Physical meaning: If y = s(t) is position and x = t is time, then dy/dx = ds/dt is velocity. Second derivative d²y/dx² = dv/dt is acceleration. JEE frequently uses this in kinematics problems.

Standard Derivative Formulas

Function	Derivative
xⁿ	nxⁿ⁻¹
sin x	cos x
cos x	-sin x
tan x	sec²x
eˣ	eˣ
ln x	1/x
aˣ	aˣ ln a
logₐ x	1/(x ln a)
sin⁻¹ x	1/√(1-x²)
cos⁻¹ x	-1/√(1-x²)
tan⁻¹ x	1/(1+x²)

Product and Quotient Rules

Product Rule: d/dx(uv) = u’v + uv’ Example: d/dx(x² sin x) = 2x sin x + x² cos x

Quotient Rule: d/dx(u/v) = (u’v - uv’)/v² Example: d/dx(x/ln x) = [(1)(ln x) - x(1/x)]/(ln x)² = (ln x - 1)/(ln x)²

Chain Rule (Composite Functions)

If y = f(g(x)), then dy/dx = f’(g(x)) · g’(x)

Example: d/dx(sin³x) = d/dx[(sin x)³] = 3(sin x)² · cos x

For a general composite: d/dx[f₁(f₂(…fₙ(x)))] = f₁’(f₂(…)) · f₂’(f₃(…)) · … · fₙ’(x)

Implicit Differentiation

When equation is not solved for y (e.g., x² + xy + y² = 5), differentiate both sides w.r.t. x, treating y as function of x:

2x + y + x(dy/dx) + 2y(dy/dx) = 0 dy/dx(x + 2y) = -2x - y dy/dx = -(2x + y)/(x + 2y)

Inverse function rule: If y = f⁻¹(x), then f’(f⁻¹(x)) · (dy/dx) = 1, or equivalently dy/dx = 1/f’(y)

Logarithmic Differentiation

For products or quotients with many factors, take ln of both sides first:

If y = xˣ, then ln y = x ln x Differentiating: (1/y)(dy/dx) = ln x + 1 dy/dx = xˣ(ln x + 1)

This handles cases like y = (x² + 1)³/(x + 1)⁴ or y = x¹/³ · (x² + 1)²/√(x + 3)

Parametric Differentiation

If x = f(t) and y = g(t), then: $$frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g’(t)}{f’(t)}$$

Second derivative: $$frac{d^2y}{dx^2} = \frac{d}{dx}left(frac{dy}{dx}right) = \frac{d}{dt}left(frac{dy}{dx}right) / \frac{dx}{dt}$$

⚡ JEE pattern: Parametric forms often appear in questions involving curves like cycloids, astroids, and epicycloids.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Differentiation — Comprehensive Notes

Limit Definition and Differentiability

A function f(x) is differentiable at x = a if: $$lim_{x \to a} \frac{f(x) - f(a)}{x - a}$$

exists (equivalently, left-hand limit = right-hand limit). If f is differentiable at a point, it must be continuous there (but converse is false — continuity ≠ differentiability).

Counterexample: f(x) = |x| is continuous at x = 0 but not differentiable there. Left-hand derivative = -1, right-hand derivative = +1.

Cusp (f(x) = |x|³ at origin): derivative exists despite corner — slopes match from both sides.

Vertical tangent (f(x) = x^(1/3)): derivative is infinite at x = 0; tangent is vertical line x = 0.

Mean Value Theorem and Rolle’s Theorem

Rolle’s Theorem: If f is continuous on [a,b], differentiable on (a,b), and f(a) = f(b), then there exists c ∈ (a,b) such that f’(c) = 0.

Mean Value Theorem (Lagrange): If f is continuous on [a,b] and differentiable on (a,b), then there exists c ∈ (a,b) such that: $$f’(c) = \frac{f(b) - f(a)}{b - a}$$

This is geometrically the slope of chord AB equals slope of tangent at some c between a and b.

Cauchy’s Mean Value Theorem: If f, g are continuous on [a,b] and differentiable on (a,b) with g’(x) ≠ 0, then: $$\frac{f’(c)}{g’(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}$$

JEE Advanced frequently uses MVT in proving inequalities.

Higher Order Derivatives

The nth derivative of a function is denoted dⁿy/dxⁿ or f⁽ⁿ⁾(x).

Leibniz Rule for nth derivative of product: $$(fg)^{(n)} = \sum_{k=0}^{n} \binom{n}{k} f^{(k)} g^{(n-k)}$$

Standard nth derivatives to remember:

• dⁿ/dxⁿ(sin x) = sin(x + nπ/2)
• dⁿ/dxⁿ(cos x) = cos(x + nπ/2)
• dⁿ/dxⁿ(eᵃˣ) = aⁿeᵃˣ
• dⁿ/dxⁿ(xⁿ) = n! (for positive integer n)

Taylor and Maclaurin Series

If f is infinitely differentiable at x = a: $$f(x) = f(a) + f’(a)(x-a) + \frac{f”(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + …$$

Maclaurin series (a = 0): $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + …$$ $$sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - …$$ $$cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - …$$

JEE Application: Using first 2-3 terms of Taylor series to approximate values or evaluate limits that are otherwise difficult (e.g., limit as x→0 of (sin x - x + x³/6)/x⁵).

L’Hôpital’s Rule

For indeterminate forms 0/0 or ∞/∞: $$lim_{x \to a} \frac{f(x)}{g(x)} = lim_{x \to a} \frac{f’(x)}{g’(x)}$$

Can be applied repeatedly if conditions hold. Other forms (0·∞, ∞-∞, 0⁰, ∞⁰, 1∞) must first be converted to 0/0 or ∞/∞ form using logarithms or algebra.

Caution: Always verify the indeterminate form before applying. Each application requires the limit to still be indeterminate.

Derivative as Rate Measure

dy/dx represents rate of change of y with respect to x. If two related quantities change with time: $$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$$

Examples in JEE:

• A ladder sliding down wall: x² + y² = L² → 2x(dx/dt) + 2y(dy/dt) = 0
• Expanding sphere: V = (4/3)πr³ → dV/dt = 4πr²(dr/dt) = surface area × rate of change of radius
• Connected pulleys: velocity relationship from string length constraint

Curvature and Radius of Curvature

Radius of curvature at point on y = f(x): $$R = \frac{[1 + (dy/dx)^2]^{3/2}}{|d^2y/dx^2|}$$

For parametric form (x(t), y(t)): $$R = \frac{[(\dot{x})^2 + (\dot{y})^2]^{3/2}}{|\dot{x}\ddot{y} - \dot{y}\ddot{x}|}$$

where dots denote derivatives w.r.t. parameter t.

Successive Differentiations and Jacobians

For functions of multiple variables u(x,y), v(x,y):

• First derivative: partial derivatives ∂u/∂x, ∂u/∂y
• Jacobian: J = ∂(u,v)/∂(x,y) = |∂u/∂x ∂u/∂y; ∂v/∂x ∂v/∂y|

Jacobians appear in change of variables for multiple integrals and in implicit function theory.

⚡ Exam tips for JEE Advanced:

1. Chain rule is the most tested concept — if answer looks messy, reapply chain rule
2. Implicit differentiation is mandatory for questions with relations like x³ + y³ = 3axy
3. For function of function of function, differentiate step by step
4. Parametric differentiation: always compute dy/dt and dx/dt separately before dividing
5. MVT questions often appear as “prove that f(b) > f(a) when…” — use the slope inequality
6. For nth derivative problems, try to find pattern from first few derivatives
7. Taylor series beyond first 2 terms rarely needed; usually first correction term suffices
8. L’Hôpital’s rule requires 0/0 or ∞/∞ form — check before applying

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