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CBSE Class 12 Maths

Continuity and Differentiability

Each concept is a short lesson, then questions until you get three right.

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17 concepts, a short lesson each

  • Continuity0/3
  • Discontinuity0/1
  • The derivative0/1
  • Composite and inverse functions0/5
  • Other ways to differentiate0/4
  • Higher order derivatives0/3

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Continuity at a point

ff is continuous at aa when f(a)f(a) is defined, the limit exists, and lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a)

Start

Continuity0 of 3

  1. Continuity at a pointff is continuous at aa when f(a)f(a) is defined, the limit exists, and lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a)Start
  2. Continuity from the left and rightLeft-continuous: lim⁡x→a−f(x)=f(a)\lim_{x \to a^-} f(x) = f(a); continuous means both sidesStart
  3. Finding constants for continuitySet each one-sided limit equal to the value at the joining point and solve for the constantsStart

Discontinuity0 of 1

  1. Continuity on an intervalPolynomials, sin⁡\sin, cos⁡\cos and exe^x are continuous everywhere; check rational functions, [x][x] and the joins of piecesStart

The derivative0 of 1

  1. Differentiability and continuityLeft and right derivatives must agree; differentiable implies continuous, but |x| is continuous with a corner at 0Start

Composite and inverse functions0 of 5

  1. The chain ruleIf y=f(u)y = f(u) and u=g(x)u = g(x), then dydx=dydu⋅dudx\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}: differentiate the outside, keep the inside, multiply by the inside's derivativeStart
  2. Chain rule with exponentials and logarithmsddxeu=euu′\dfrac{d}{dx}e^{u} = e^{u}u', ddxau=aulog⁡a⋅u′\dfrac{d}{dx}a^{u} = a^{u}\log a\cdot u', ddxlog⁡u=u′u\dfrac{d}{dx}\log u = \dfrac{u'}{u}; simplify logs of products firstStart
  3. Derivative of an inverse functionIf y=f(x)y = f(x) has inverse x=g(y)x = g(y), then g′(y)=1f′(x)g'(y) = \dfrac{1}{f'(x)} at the matching point: the slopes of a graph and its mirror in y=xy = x are reciprocalsStart
  4. Derivatives of inverse trigonometric functionsddxsin⁡−1x=11−x2\dfrac{d}{dx}\sin^{-1}x = \dfrac{1}{\sqrt{1 - x^2}}, ddxtan⁡−1x=11+x2\dfrac{d}{dx}\tan^{-1}x = \dfrac{1}{1 + x^2}, ddxsec⁡−1x=1∣x∣x2−1\dfrac{d}{dx}\sec^{-1}x = \dfrac{1}{|x|\sqrt{x^2 - 1}}; with the chain rule for sin⁡−1u\sin^{-1}uStart
  5. Simplifying by substitution firstPut x=tan⁡θx = \tan\theta, sin⁡θ\sin\theta or cos⁡θ\cos\theta so that tan⁡−12x1−x2\tan^{-1}\dfrac{2x}{1 - x^2} becomes 2θ=2tan⁡−1x2\theta = 2\tan^{-1}x, then differentiate the simple formStart

Other ways to differentiate0 of 4

  1. Logarithmic differentiationTake log⁡\log of both sides to turn products, quotients and powers like xxx^{x} into sums, differentiate, then multiply back by yyStart
  2. Implicit differentiationDifferentiate every term in xx, writing ddx(yn)=nyn−1dydx\dfrac{d}{dx}(y^n) = ny^{n-1}\dfrac{dy}{dx}, then collect the dydx\dfrac{dy}{dx} terms and solveStart
  3. Parametric functionsIf x=f(t)x = f(t) and y=g(t)y = g(t), then dydx=dy/dtdx/dt\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}, the slope of the curve traced as tt variesStart
  4. Differentiating one function with respect to anotherThe derivative of uu with respect to vv is dudv=du/dxdv/dx\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx}Start

Higher order derivatives0 of 3

  1. Second and higher derivativesDifferentiate dydx\dfrac{dy}{dx} again for d2ydx2\dfrac{d^2y}{dx^2}; for eaxe^{ax}, sin⁡ax\sin ax and similar, spot the nnth derivative's patternStart
  2. Second derivative of parametric and implicit functionsFor x=f(t)x = f(t), y=g(t)y = g(t): d2ydx2=ddt(dydx)/dxdt\dfrac{d^2y}{dx^2} = \dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)\Big/\dfrac{dx}{dt}, not the ratio of second derivativesStart
  3. Showing a relation between derivativesTo show (1−x2)y2−xy1+m2y=0(1 - x^2)y_2 - xy_1 + m^2y = 0: find y1y_1, clear roots or fractions, differentiate again, then substitute backStart

Practise Continuity and Differentiability

Once the concepts click, try the questions your exam asks.