How do you differentiate an inverse function?

Write x in terms of y, find \frac{dx}{dy}, then \frac{dy}{dx}=\frac{1}{dx/dy}.

Why is the positive root taken for \cos y when y=\arcsin x?

y lies in \left[-\frac\pi2,\frac\pi2\right], where \cos y\geq0.

Differentiating inverse trigonometric and hyperbolic functionsEdexcel International A Level Further Maths: Flashcards

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Question

$\frac{d}{dx}\arcsin x$?

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$\frac{d}{dx}\arcsin x$ ?: $\frac{1}{\sqrt{1-x^2}}$
$\frac{d}{dx}\arccos x$ ?: $-\frac{1}{\sqrt{1-x^2}}$
$\frac{d}{dx}\arctan x$ ?: $\frac{1}{1+x^2}$
$\frac{d}{dx}\operatorname{arsinh}x$ ?: $\frac{1}{\sqrt{x^2+1}}$
$\frac{d}{dx}\operatorname{arcosh}x$ ?: $\frac{1}{\sqrt{x^2-1}}$ , for $x>1$
$\frac{d}{dx}\operatorname{artanh}x$ ?: $\frac{1}{1-x^2}$ , for $|x|<1$
How do you differentiate an inverse function?: Write $x$ in terms of $y$ , find $\frac{dx}{dy}$ , then $\frac{dy}{dx}=\frac{1}{dx/dy}$ .
$\frac{d}{dx}\arctan(x^2)$ ?: $\frac{2x}{1+x^4}$
$\frac{d}{dx}\left(\frac12\operatorname{artanh}x^2\right)$ ?: $\frac{x}{1-x^4}$
$\frac{d}{dx}\left(\arcsin x+x\sqrt{1-x^2}\right)$ ?: $2\sqrt{1-x^2}$
$\frac{d}{dx}\left(x\operatorname{arsinh}x-\sqrt{1+x^2}\right)$ ?: $\operatorname{arsinh}x$
$\operatorname{arsinh}x$ in logarithmic form?: $\ln\left(x+\sqrt{x^2+1}\right)$
Why is the positive root taken for $\cos y$ when $y=\arcsin x$ ?: $y$ lies in $\left[-\frac\pi2,\frac\pi2\right]$ , where $\cos y\geq0$ .

Exam questions on Differentiating inverse trigonometric and hyperbolic functions

Let $y=\operatorname{arsinh}x$ , so that $\sinh y=x$ .
Show that $\frac{dy}{dx}=\frac{1}{\sqrt{1+x^2}}$ .2 marks
The function $f$ is defined by $f(x)=x\arctan x$ .
Show that $f$ has exactly one stationary point.2 marks
The curve $C$ has equation $y=\arcsin x+x\sqrt{1-x^2}$ for $-1<x<1$ .
Show that $\frac{dy}{dx}=2\sqrt{1-x^2}$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).