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MH HSC 12 Maths

Differentiation

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

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12 concepts, a short lesson each

  • Composite and inverse functions0/5
  • Other ways to differentiate0/4
  • Higher order derivatives0/3

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The chain rule

If 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 derivative

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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 Differentiation

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

Textbook exercisesEvery exercise, with model answersPast board questionsAsked from this chapter, with solutions