# How do you find the derivative of y=cos^-1(e^(2x))?

Oct 22, 2016

Let's start by determining the derivative of $y = {\cos}^{-} 1 \left(x\right)$.

$y = {\cos}^{-} 1 \left(x\right)$

$\cos y = x$

Through implicit differentiation, you should have:

$- \sin y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sin} y$

Since $\sin y = \sqrt{1 - {\cos}^{2} y}$, we can rewrite as:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sqrt{1 - {\cos}^{2} y}}$

Since $x = \cos y$, we can substitute:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sqrt{1 - {x}^{2}}}$

Now, let's differentiate ${e}^{2 x}$. The derivative of any function of the form ${e}^{f} \left(x\right)$ is $f ' \left(x\right) \times {e}^{f} \left(x\right)$. So, the derivative is $2 {e}^{2 x}$.

By the chain rule, letting $y = {\cos}^{-} 1 \left(u\right)$ and $u = {e}^{2 x}$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sqrt{1 - {u}^{2}}} \times 2 {e}^{2 x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 {e}^{2 x}}{\sqrt{1 - {e}^{2 x}}}$

Hopefully this helps!