# How do you find the derivative of arcsin e^x?

Jun 23, 2016

${e}^{x} / \sqrt{1 - {e}^{2 x}}$

#### Explanation:

There are two methods:

Using the pre-memorized arcsine derivative:

You may already know that the derivative of arcsine is:

$\frac{d}{\mathrm{dx}} \arcsin \left(x\right) = \frac{1}{\sqrt{1 - {x}^{2}}}$

We can apply the chain rule to this for $\arcsin \left({e}^{x}\right)$:

$\frac{d}{\mathrm{dx}} \arcsin \left({e}^{x}\right) = \frac{1}{\sqrt{1 - {\left({e}^{x}\right)}^{2}}} \cdot \frac{d}{\mathrm{dx}} {e}^{x}$

$= {e}^{x} / \sqrt{1 - {e}^{2 x}}$

Without knowing the arcsine derivative:

Let

$y = \arcsin \left({e}^{x}\right)$

Thus:

$\sin \left(y\right) = {e}^{x}$

Differentiate both sides (the chain rule will be used on the left-hand side!):

${y}^{'} \cdot \cos \left(y\right) = {e}^{x}$

${y}^{'} = {e}^{x} / \cos \left(y\right)$

Note that we should express $\cos \left(y\right)$ in terms of $\sin \left(y\right)$, since we know that $\sin \left(y\right) = {e}^{x}$.

We know that

${\sin}^{2} \left(y\right) + {\cos}^{2} \left(y\right) = 1 \text{ "=>" } \cos \left(y\right) = \sqrt{1 - {\sin}^{2} \left(y\right)}$

${y}^{'} = {e}^{x} / \sqrt{1 - {\sin}^{2} \left(y\right)}$

And since $\sin \left(y\right) = {e}^{x}$:

${y}^{'} = {e}^{x} / \sqrt{1 - {\left({e}^{x}\right)}^{2}} = {e}^{x} / \sqrt{1 - {e}^{2 x}}$