Derivative of the $\arctan(x)$

Question

I'm wonder how this trigonometric function has a rational function derivative!!

Derivative of inverse is equal to inverse of derivative, isn't it? So why it is $1/(1+x^2)$ instead of $1/(1+\tan^2(x))$?

Answer 1

The derivative of the inverse is not equal to the inverse of the derivative. Let $f(x)$ be a function and let $f^{-1}(x)$ denote the inverse function. Then $f(f^{-1}(x))=x$. Take the derivative of both sides and use the chain rule: $$ f'(f^{-1}(x))\cdot (f^{-1}(x))' = 1. $$ Therefore the derivative of the inverse function is $$ \frac{1}{f'(f^{-1}(x))}. $$ In this case $f(x)=\tan x$ and $f'(x)=(\sec x)^2=1+(\tan x)^2$. Thus the derivative of $\tan^{-1}x$ is $$ \frac{1}{1+(\tan(\tan^{-1}x))^2}=\frac{1}{1+x^2}. $$

Answer 2

Knowing $arctg(tg(x))=x$

By chain rule we get:

$$[arctg(tg(x))]'=x'=1$$

So

$$tg'(x) \times arctg'(tg(x))=1$$

Now, set $tg(x)=y$:

$$(1+y^2) \times arctg'(y)=1$$

Hence

$$arctg'(y)=\frac{1}{1+y^2}$$

and it's not trigonometric!

The derivative of inverse is equal to the inverse of derivative in another point $(y)$ and it cancel out with the original function.