Chain Rule and Standard Derivatives

MATTechniques of Differentiation

The chain rule differentiates composite functions by multiplying the derivative of the outer function by the derivative of the inner. Combined with the standard derivative table (powers, trig, inverse-trig, exponential, log), it handles almost every differentiation in ISC.

A high-yield twist is simplifying inverse-trig arguments using substitutions like $x=\tan\theta$ before differentiating.

Chain rule

\dfrac{d}{dx}\,f(g(x)) = f'(g(x))\cdot g'(x)

Outer derivative evaluated at the inner function, times the inner derivative; extends to several nested layers.

Inverse-trig derivatives

\dfrac{d}{dx}\tan^{-1}x = \dfrac{1}{1+x^2}, \quad \dfrac{d}{dx}\sin^{-1}x = \dfrac{1}{\sqrt{1-x^2}}

\sin^{-1}

derivative valid on

-1<x<1

;

\tan^{-1}

for all real

x

. Apply the chain rule when the argument is a function of

x

Exponential and logarithmic derivatives

\dfrac{d}{dx}e^{u} = e^{u}\dfrac{du}{dx}, \qquad \dfrac{d}{dx}\log u = \dfrac{1}{u}\dfrac{du}{dx}

u=u(x)

u>0

for

\log u

; e.g. for

y=e^{\sin^{-1}x}

\dfrac{dy}{dx}=e^{\sin^{-1}x}\cdot\dfrac{1}{\sqrt{1-x^2}}

Standard simplification substitution

\tan^{-1}\dfrac{2x}{1-x^2} = 2\tan^{-1}x \quad (-1<x<1)

Put

x=\tan\theta

so the argument becomes

\tan 2\theta

; similarly

\sin^{-1}\dfrac{2x}{1+x^2}=2\tan^{-1}x

for

-1\le x\le 1

Differentiate from the outermost function inward, multiplying derivatives of every layer.
Before differentiating a messy inverse-trig expression, simplify it using $x=\tan\theta$ , $x=\sin\theta$ , or $x=a\tan\theta$ to reduce it to a multiple of $\theta$ .
$\tan^{-1}\dfrac{\cos x-\sin x}{\cos x+\sin x}=\dfrac{\pi}{4}-x$ , so its derivative is simply $-1$ .
$\log\tan\!\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)$ differentiates to $\sec x$ , a classic result.
For $\tan^{-1}\dfrac{\sqrt{1+x^2}-1}{x}$ , substitute $x=\tan\theta$ to get $\dfrac12\tan^{-1}x$ , whose derivative is $\dfrac{1}{2(1+x^2)}$ .
Keep track of validity intervals: the simplified linear form of an inverse-trig identity may only hold on part of the domain.
Product and quotient rules combine with the chain rule for terms like $x^2 e^{3x}$ or $\dfrac{\sin x}{1+x^2}$ .

Where the marks go

Forgetting to multiply by the inner derivative, e.g. writing $\dfrac{d}{dx}\sin(3x)=\cos(3x)$ instead of $3\cos(3x)$ .
Differentiating a complicated inverse-trig expression term-by-term instead of simplifying first, leading to long error-prone algebra.
Using the wrong sign or domain for inverse-trig identities, e.g. assuming $\tan^{-1}\dfrac{2x}{1-x^2}=2\tan^{-1}x$ holds outside $-1<x<1$ .
Confusing $\dfrac{d}{dx}\tan^{-1}x=\dfrac{1}{1+x^2}$ with $\dfrac{d}{dx}\tan x=\sec^2 x$ .

How the board asks it

Numericalthe chain rule applied to nested standard functions
Find $\dfrac{dy}{dx}$ if $y = \log\left(\sin\left(e^{x^2}\right)\right)$ .
Numericalthe substitution $x=\tan\theta$ to reduce an inverse-trig argument
Differentiate $\tan^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)$ with respect to $x$ , after simplifying it using a suitable substitution.
Predict the productthe product rule combined with the chain rule
If $y = x^2\, e^{3x}\sin x$ , find $\dfrac{dy}{dx}$ .
Derive / provethe classic result $\log\tan\left(\tfrac{\pi}{4}+\tfrac{x}{2}\right)$
If $y = \log\tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)$ , prove that $\dfrac{dy}{dx} = \sec x$ .

Practise this topic

PreviousDifferentiability and First Principles NextImplicit Differentiation

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.