How can I find the derivative of $y = \sin(\arctan x) + \tan(\arcsin x)$? My question is, how can I solve the following derivative question?
$$y = \sin(\arctan x) + \tan(\arcsin x)$$
Thanks in advance

First you want to simplify $\tan (\arcsin x)$ and $\sin(\arctan x)$. Let me show
you the latter one.
Suppose $\theta=\arctan x$. Then by definition $-\pi/2<\theta<\pi/2$ and $\tan\theta=x$.
Since $\sec \theta>0$ (because $-\pi/2<\theta<\pi/2$), and $$\sec^2\theta=1+\tan^2 \theta =1+x^2,$$ you get
$\sec \theta=\sqrt{1+x^2}$, so $\cos \theta =1/\sqrt{1+x^2},$ and 
$$\sin\theta=\tan\theta \cos\theta =\frac{x}{\sqrt{1+x^2}},$$
i.e. $\sin(\arctan x)=\frac{x}{\sqrt{1+x^2}}$. 
Similarly, you can prove $\tan (\arcsin x)=\frac{x}{\sqrt{1-x^2}}.$
Using quotient rule, you can prove that the derivative of $\frac{x}{\sqrt{1+x^2}}$
and $\frac{x}{\sqrt{1-x^2}}$ are 
$$\frac{1}{(1+x^2)^{3/2}} \mbox{ and }\frac{1}{(1-x^2)^{3/2}} $$
respectively. Hence the derivative of  your function is $$\frac{1}{(1+x^2)^{3/2}}+\frac{1}{(1-x^2)^{3/2}}. $$
${\bf Edit.}$ Or you can use chain rule to get:
$$\cos(\arctan x)\frac{1}{1+x^2}+\sec^2(\arcsin x)\frac{1}{\sqrt{1-x^2}}$$
Now simplify it.
