Find the derivative of $G(x) = \tan^{-1}(x^2 + 1)$

Question

Find the derivative of $G(x) = \tan^{-1}(x^2 + 1)$

I’m having trouble with derivatives. Could someone please help me with this? I’m not sure exactly how to go on in solving this problem.

Answer 1

HINT

Let use the chain rule

$$g(x)=f(h(x))\implies g'(x)=f'(h(x))\cdot h'(x)$$

that is

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

with $$f(x)=x^2+1\implies f'(x)=2x$$

Answer 2

$$\frac{d}{dx}[ \tan^{-1} (x^2 + 1)] = \frac{1}{1+(x^2+1)^2}\times \frac{d}{dx}[x^2 + 1] = \frac{2x}{x^4 + 2x^2 + 2}$$