What is the derivative of arccot(e^2x)? I did this problem by switching the expression to arctan[e^(-2x)] then got (-2e^-2x)/(e^-4x + 1) but my book says (-2e^2x)/(1 + e^4x).
What am I doing wrong?
Let $\displaystyle y=\cot^{-1}(e^{2x})$, then $\displaystyle \cot(y)=e^{2x}, \tan(y)=e^{-2x}$
Now differentiate both sides wrt x:
$\displaystyle \sec^2(y)\frac{dy}{dx}=-2e^{-2x}$ and so $\displaystyle \frac{dy}{dx}=\frac{-2e^{-2x}}{\sec^2(y)}$
Using $\displaystyle \sec^2(y)=1+\tan^2(y) = 1+e^{-4x}$ we have
$\displaystyle \frac{dy}{dx}=\frac{-2e^{-2x}}{1+e^{-4x}}$
Now multiply top and bottom by e^{4x} to get $\displaystyle \frac{-2e^{2x}}{e^{4x}+1}$.