Prove $(\tan^{-1}{x})' = \frac{1}{1+x^2}$ Prove $(\tan^{-1}{x})' = \frac{1}{1+x^2}$
How should I proceed? I tried: 

Then another which I think is wrong: 

What should I be doing? 
UPDATE
In my lecture notes ... maybe a typo: 

 A: I guess that you want to prove that $(\tan^{-1} x)'=1/(1+x^2)$.
$$ (\tan^{-1} x)'=\frac{1}{\tan' y}$$
where $\tan y=x$. Therefore
$$ (\tan^{-1} x)'=\frac{1}{1+\tan^2 y}=\frac{1}{1+x^2}$$
A: The answer above is also correct, but for clarity here it goes
Let $$y=\tan^{-1} x \Rightarrow \tan(y) = x$$
Differentiate both sides:
Left side will be $\sec^2(y) \dfrac{dy}{dx}$, Right side will be $1$
But using the trigonometric identity $\sec^{2}(y) = 1 + \tan^{2}(y)$
$$\begin{align}
(1+\tan^{2}(y))&\frac{dy}{dx} = 1 \\
\Rightarrow& \frac{dy}{dx} = \frac{1}{1+\tan^{2}(y)}
\end{align}$$
Since we began with $y=\tan^{-1} x \Rightarrow \tan(y) = x$
$$\frac{dy}{dx} = \frac{1}{1+x^2}$$
But $\dfrac{dy}{dx} = (\tan^{-1} x)'$, Therefore $$(\tan^{-1}x)' = \frac{1}{1+x^2}$$
A: It should be $(\tan^{-1}x)'$ (or arc$\tan x$). Then use $$y=\tan^{-1} x$$
$$\Rightarrow \tan y= x$$ Now differentiate both sides w.r.t. $x$ to get $sec^2 y. y'=1$, or $(1+\tan^2 y).y'=1$, or $(1+x^2).y'=1$
A: Has your class covered integration?  It may be easier to prove that $\int\frac{dx}{1+x^2}=\tan^{-1}x+C$.  To clear up the denominator, try the substitution
$x=\tan\theta,dx=\sec^2\theta d\theta$
$\int\frac{dx}{1+x^2}=\int\frac{\sec^2\theta d\theta}{1+\tan^2\theta}=\int\frac{\sec^2\theta d\theta}{\sec^2\theta}=\int d\theta=\theta+C$
Since $x=\tan\theta, \theta=\tan^{-1}x$
