# What is the derivative of this function e ^ sin (2x)?

May 7, 2018

$\frac{d}{\mathrm{dx}} \left({e}^{\sin} \left(2 x\right)\right) = 2 {e}^{\sin} \left(2 x\right) \cos \left(2 x\right)$

#### Explanation:

Using the chain rule with $y = \sin t$ and $t = 2 x$ we have:

${e}^{\sin} \left(2 x\right) = {e}^{y}$

$\frac{d}{\mathrm{dx}} \left({e}^{\sin} \left(2 x\right)\right) = \frac{d}{\mathrm{dy}} \left({e}^{y}\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

and similary:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \sin t = \frac{d}{\mathrm{dt}} \sin t \cdot \frac{\mathrm{dt}}{\mathrm{dx}} = \cos t \cdot \frac{\mathrm{dt}}{\mathrm{dx}}$

and finally:

$\frac{\mathrm{dt}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(2 x\right) = 2$

Putting it together:

$\frac{d}{\mathrm{dx}} \left({e}^{\sin} \left(2 x\right)\right) = 2 {e}^{\sin} \left(2 x\right) \cos \left(2 x\right)$

May 7, 2018

$2 {e}^{\sin 2 x} \cos 2 x$

#### Explanation:

Given: ${e}^{\sin 2 x}$.

Apply the chain rule, which states that,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Let $u = \sin 2 x$, then we must find $\frac{\mathrm{du}}{\mathrm{dx}}$.

Again, use the chain rule. Let $z = 2 x , \therefore \frac{\mathrm{dz}}{\mathrm{dx}} = 2$.

Then $y = \sin z , \frac{\mathrm{dy}}{\mathrm{dz}} = \cos z$.

Combine to get: $\cos z \cdot 2 = 2 \cos z$.

Substitute back $z = 2 x$ to get:

$= 2 \cos \left(2 x\right)$

Now, we go back to the original derivative.

From here, $y = {e}^{u} , \therefore \frac{\mathrm{dy}}{\mathrm{du}} = {e}^{u}$, and now we can combine our results from the previous findings, where $\frac{\mathrm{du}}{\mathrm{dx}} = 2 \cos 2 x$.

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{u} \cdot 2 \cos 2 x$

$= 2 {e}^{u} \cos 2 x$

Substitute back $u = \sin 2 x$ to get:

$= 2 {e}^{\sin 2 x} \cos 2 x$