So, how do I show that $1+x^2=\sec^2(\arctan(x))$?/How to find derivative of $\arctan(x)$? So I figured I can use the chain rule to do this:
$g\prime(x)=\frac{1}{f^\prime(g(x))}$
So that
$(\arctan(x))\prime = \frac{1}{\left[\sec^2(\arctan(x)){}\right]^\prime}$
But this book tells me that 
$(\arctan(x))\prime = \frac{1}{x^2+1}$

So, how do I show that $1+x^2=\sec^2(\arctan(x))$?

 A: Use the trigonometric identity $1+\tan^2 x=\sec^2 x$.
A: The following geometric approach can be useful.
First suppose that $x \gt 0$. Draw a right-angled triangle $ABC$, right-angled at $C$. Look at $\angle A$. Let the length of $BC$ be $x$ (put an $x$ beside leg $BC$). Let $AC$ have length $1$. Then $\tan A=x$, so $A=\arctan x$. The hypotenuse has length $\sqrt{1+x^2}$, so $\sec A=\frac{1}{\cos A}=\frac{\sqrt{1+x^2}}{1}$, and therefore $\sec^2 A=1+x^2$.
If $x=0$ the result is trivial.
If $x \lt 0$, note that $\arctan x=-\arctan |x|$.
A: A fancy, longer way:
$$(1+x^2)'=2x$$
$$[\sec^2(\arctan x)]'=2\sec(\arctan x)\cdot\tan(\arctan x)\cdot\sec(\arctan x)\cdot\frac{1}{1+x^2}=$$
$$=2\sec^2(\arctan x)\frac{x}{1+x^2}=2(1+\tan^2(\arctan x))\frac{x}{1+x^2}=$$
$$=2(1+x^2)\frac{x}{1+x^2}=2x$$
Thus, we got:
$$(1+x^2)'=[\sec^2(\arctan x)^2]'\Longleftrightarrow 1+x^2=\sec^2(\arctan x)+C\,\,,\,C=\text{ a constant}$$
To find $\,C\,$ now just evaluate above for $\,x=0\,$ , say...
A: Note that $sec^{2}u=1+tan^{2}u$. Let $u=arctan(x)$. That should take care of the equality you want. The usual way to get the derivative of $arctan(x)$, at least from my experience, is to let $y=arctan(x)$ and conclude that $tan(y)=x$, by definition of the $arctan$ function. Now use implicit differentiation. You should get $\frac{dy}{dx}=\frac{1}{sec^{2}y}$.Substitute for $sec^{2}y$ to get $\frac{dy}{dx}=\frac{1}{1+tan^{2}y}$. But $tan(y)=x$, and you get the desired result. 
