Find $\left(\arctan\left({1+x^2}\right)\right)'$

Question

I want to find the derivative of

$$\left(\arctan\left({1+x^2}\right)\right)'$$

From the derivatives table I see that

$$\arctan{u}=\frac 1 {1+u^2}$$

Therefore it is intuitive for me to replace $1+x^2$ (which is the argument of the function above) with the $u$ of the right side:

$$\left(\arctan\left({1+x^2}\right)\right)'=\frac{1}{1+(1+x^2)^2}$$

This is not the right solution (the numerator should be equal to $2x$). I understand that $(1+x^2)'=2x$ though I don't know why we have to put $2x$ at the top. Any hints?

Answer 1

Hint:

$f(g(x))'=f'(g(x))g'(x)$

Answer 2

Hint:

use the chain rule of derivatives.

Define $f(x)=\arctan(x)$ and $g(x)=1+x^2$. Then, $$\arctan(1+x^2)=f(g(x))=(f\circ g)(x)$$

and you can use the fact that $(f\circ g)'(x) = (f\circ g')(x)\cdot g'(x)$

Answer 3

$y= arc tan (1+x^2)$

$\tan y =1+x^2$

Taking derivative of both sides we get:

$y'(1+\tan^2 y)= 2x$

$(arc tan (1+x^2))'=y'=\frac{2x}{1+(1+x^2)^2}$