Does the derivative of $\arctan(x) = \frac{1}{1+x^2} = \cos^2(y)$, where $x = \tan(y)$?

Question

I know the derivative of $\arctan(x) = \frac{1}{1+x^2}$ and the derivative of $\tan(x) = \frac{1}{\cos^2(x)}$. But, I recently read somewhere that the derivative of the inverse function is the reciprocal of the derivative. So the derivative of $\arctan(x) = \cos^2(y)$, where $x = \tan(y)$. Does this hold? I played around with some numbers and it seems it works.

Answer 1

If you differentiate a function depending on $x$, the answer should also be a function depending on $x$. You have:\begin{align}\arctan'(x)&=\frac1{\tan'(\arctan x)}\\&=\frac1{^\frac1{\cos^2(\arctan x)}}\\&=\frac1{1+\tan^2(\arctan x)}\text{ (because $\frac1{\cos^2\theta}=1+\tan^2\theta$)}\\&=\frac1{1+x^2}.\end{align}In the general case, we have$$(f^{-1})'(x)=\frac1{f'\bigl(f^{-1}(x)\bigr)}.$$

Answer 2

If $x = \tan(y)$, then $$\cos^2(y) = \frac1{\sec^2(y)} = \frac1{1+\tan^2(y)} = \frac1{1+x^2}.$$

Answer 3

You are right, $$(\arctan(x))'=\frac1{1+x^2}$$ is true and if $x=\tan(y)$,

$$\frac1{1+x^2}=\cos^2(y)$$ as well.

You can also write

$$(\arctan(x))'=\cos^2(\arctan(x)).$$

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Now if you write

$$y'=\cos^2(y),$$ this is an ordinary differential equation, which is autonomous, and is satisfied by

$$y=\arctan(x+c).$$