We have to show that $(\arctan(x))' = \frac{1}{1+x^2}$, derived with the chain rule.
The hint given is that we should start with deriving $\tan(\arctan(x)) = x$; I am not sure though how this is helpful, since the derivative of $\arctan(x)$ is what we are looking for yet..
Let $y = \arctan x.$
Then $x = \tan\Big(\arctan x\Big)= \tan y$
Then $$\dfrac{dx}{dy}=\sec^2y = 1 +\tan^2 y$$
Since we have $x = \tan y$, $$\dfrac{dx}{dy}= 1 + (\tan y)^2 = 1+x^2 $$ Hence, it follows that: $$\frac{dy}{dx}=\dfrac{d}{dx}\left(\arctan x\right) = \dfrac{1}{1+x^2 }$$
You are given $\tan(\tan^{-1} x) = x$. Take a derivative of that to get $\displaystyle \sec^2(\tan^{-1} x) \cdot \frac{d}{dx} \left(\tan^{-1} x\right) = 1$.
Now, consider a right triangle with adjacent length $1$ and opposite length $x$, so that the triangle has an angle $\theta$ such that $\tan \theta = x$. Now we take $\displaystyle \sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{1}{\sqrt{x^2+1}}} = \sqrt{x^2+1}$. So, $\sec^2 (\tan^{-1} x) = \sec^2 \theta = x^2+1$. Plugging this back into the original equation, we get $$\frac{d}{dx} \left(\tan^{-1} x\right) = \frac{1}{x^2+1}.$$
That's a theta, btw, not a phi. This is a theta: $\theta$. And this is a phi: $\phi$.
– 2012ssohn Oct 29 '13 at 21:10$(f\circ g)' = (f' \circ g) \cdot g'$
You are looking for $g'$, so just solve for it and substitute the functions given.
