# Question #321de

Oct 16, 2017

${e}^{\sin} x \left({\cos}^{2} x\right) - {e}^{\sin} x \left(\sin x\right)$

#### Explanation:

We know that the first derivative is ${e}^{\sin} x \left(\cos x\right)$

To get the second derivative we apply the product rule:

Where $f ' \left(g\right) + g ' \left(f\right)$

Let's say that $f$ is our ${e}^{\sin} x$ and $g$ is $\cos x$

We take the derivative of $f$ and multiply it by $g$ and add the derivative of $g$ multiplied by $f$

${e}^{\sin} x \left(\cos x\right) \left(\cos x\right) + \left(- \sin x\right) \left({e}^{\sin} x\right)$

Simplify:

${e}^{\sin} x \left({\cos}^{2} x\right) - {e}^{\sin} x \left(\sin x\right)$