Determine the derivative of $\arctan$ function Find $f'(x)$ for the function:
$f(x)= \arctan(\frac{a+x}{1-ax}))$ , $a\in R$
So this is what I've done:
$f(x) = \arctan x$
$f'(x) = \frac{1}{1+x^2}$
$x= \frac{a+x}{1-ax}$
$f'(x) =  \frac{1}{1+\frac{(a+x)^2}{(1-ax)^2}}$
$f'(x) = \frac{(1-ax)^2}{(1-ax)^2+(a+x)^2}$
Is this correct?

$$\arctan\left(\frac{x+a}{1-ax}\right)=\arctan x+\arctan a$$
(give or take a multiple of $\pi$), so
$$\frac d{dx}\arctan\left(\frac{x+a}{1-ax}\right)=\frac1{1+x^2}.$$
