# How do you find d/dx(arctan(e^2x))?

Mar 15, 2015

First of all, I am assuming that there is a mistake in the formatting, and that you mean ${e}^{2 x}$ instead of ${e}^{2} x$.

This is a composite function. Which means, a function $f$ of the form $f \left(x\right) = g \left(h \left(x\right)\right)$. In your case, $g \left(x\right) = \setminus \arctan \left(x\right)$, and $h \left(x\right) = {e}^{2 x}$. To derive such functions, you must use the chain rule, which states that, using the same notation I introduced above, $f ' \left(x\right) = g ' \left(h \left(x\right)\right) \cdot h ' \left(x\right)$. The two derivatives we need are both elementary, so I will not proof the results:
$\frac{d}{\mathrm{dx}} \setminus \arctan \left(x\right) = \frac{1}{{x}^{2} + 1}$
$\frac{d}{\mathrm{dx}} {e}^{2 x} = 2 {e}^{2 x}$

Note that for the second derivative, we actually use the chain rule one more time, because ${e}^{2 x}$ is of the form a(b(x), with $a \left(x\right) = {e}^{x}$, and $b \left(x\right) = 2 x$.

The final answer is thus, as said above,
$g ' \left(h \left(x\right)\right) \cdot h ' \left(x\right) = \frac{1}{{e}^{4 x} + 1} \cdot 2 {e}^{2 x} = \frac{2 {e}^{2 x}}{{e}^{4 x} + 1}$