Hello, pearlyc!

Is the derivative of $\displaystyle \sin[\arctan(x)] \:=\:\frac{\cos[\arctan(x)]}{1 - x^2}$ ? . . . . no

Is there anyway to further simplify this? . . . . yes The derivative is: .$\displaystyle \frac{\cos[\arctan(x)]}{1\: {\color{red}+}\: x^2} $

$\displaystyle \text{We have: }\;\cos\underbrace{[\arctan(x)]}_{\text{angle }\theta}$

Then: .$\displaystyle \theta \:=\:\arctan(x) \quad\Rightarrow\quad\tan\theta \:=\:\frac{x}{1} \:=\:\frac{opp}{adj}$

So $\displaystyle \theta$ is in this right triangle: Code:

*
____ * *
√1+x² * *
* * x
* *
* θ *
* * * * * * *
1

We have: .$\displaystyle opp = x,\;adj = 1$

From Pythagorus, we have: .$\displaystyle hyp = \sqrt{1+x^2}$

Hence: .$\displaystyle \cos\theta \:=\:\frac{1}{\sqrt{1+x^2}}$

And your answer becomes: .$\displaystyle \frac{\frac{1}{\sqrt{1+x^2}}}{1+x^2} \;=\;\frac{1}{(1+x^2)^{\frac{3}{2}}} $