Calculus Problem

There are formulas for this, but this is how I would do it:

$y=arctan(m),\quad\;\; m=cos(x)\\ tan(y)=m\\ \frac{dm}{dy}=sec^2(y)\qquad \frac{dm}{dx}=-sin(x)\\~\\ \frac{dy}{dx}=\frac{dy}{dm}\cdot \frac{dm}{dx}\\ \frac{dy}{dx}=\frac{1}{sec^2y}\cdot -sin(x)\\ \frac{dy}{dx}=\frac{-sin(x)}{sec^2(arctan(cos(x))}$

Again there are formulas but I used a series of right angled triangles to simplify the denominator. 

And I came up with  a denominator of      $1+ cos^2(x)$       

The problem with my method is that I will only guarantee if for angles in the first quadrant. 

Since it is squared it probably does not matter.

(that is assuming I made no stupid errors)

So now I have.

$\frac{dy}{dx}=\frac{-sin(x)}{1+cos^2(x)}\\ \text{Wolfram Alpha gives the asnwer as} \\ \frac{dy}{dx}=\frac{-2sin(x)}{3+cos(2x)}\\~\\ BUT\\ \text{These two answers are rearrangements of the same thing.}$

LaTex:

y=arctan(m),\quad\;\; m=cos(x)\\
tan(y)=m\\
\frac{dm}{dy}=sec^2(y)\qquad \frac{dm}{dx}=-sin(x)\\~\\
\frac{dy}{dx}=\frac{dy}{dm}\cdot \frac{dm}{dx}\\
\frac{dy}{dx}=\frac{1}{sec^2y}\cdot -sin(x)\\
\frac{dy}{dx}=\frac{-sin(x)}{sec^2(arctan(cos(x))}