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Proof that $\frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2}$ using implicit differentiation

My workings:

$y=\tan^{-1} x$,

$x= \tan y$

$\frac{d}{dx} (x) = \frac{d}{dx} \tan y$

$1 = \sec^2 y \frac{dy}{dx}$

$\frac{dy}{dx} = \frac{1}{\sec^2 y} = \cos^2 y$

How do I carry on from here?

user307640
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    you do very complicate... Since $\tan'(x)=1+\tan(x)^2=\frac{1}{\cos(x)^2}$,$$\tan(\arctan(x))=x\implies \tan'(\arctan(x))\arctan'(x)=1\implies (1+\tan(\arctan(x))^2)\arctan'(x)=1.$$ I let you finish. – Surb Sep 20 '22 at 09:22
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    What you miss is just $sec^2 y = 1 + tan^2 y$. Your steps are good, using implicit differentiation. – Abel Wong Sep 20 '22 at 10:56

2 Answers2

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Use the basic trigonometric identity $$ y'=\cos^2y=\frac{\cos^2y}1=\frac{\cos^2y}{\cos^2y+\sin^2y}=\frac1{1+\tan^2y}=\frac1{1+x^2} $$

Lutz Lehmann
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$$x=\tan y\implies x^2=\tan^2y$$$$\implies 1+x^2=1+\tan^2y=\sec^2y$$$$\implies \frac{dy}{dx}=\frac{1}{\sec^2y}=\frac{1}{1+x^2}.$$

insipidintegrator
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