# What is the second derivative for y(x)=5e^(pi x)+cos (y(x)) ?

Sep 26, 2015

$y ' ' = \frac{5 {\pi}^{2} {e}^{\pi x} \left({\left(1 + \sin y\right)}^{2} - 5 {e}^{\pi x} \cos y\right)}{1 + \sin y} ^ 3$

#### Explanation:

d/dx(y(x))=d/dx(5e^(pix)+cos(y(x))

$y ' = 5 \pi {e}^{\pi x} - \sin y \cdot y '$

$y ' = \frac{5 \pi {e}^{\pi x}}{1 + \sin y}$

$\frac{d}{\mathrm{dx}} \left(y '\right) = \frac{5 {\pi}^{2} {e}^{\pi x} \cdot \left(1 + \sin y\right) - 5 \pi {e}^{\pi x} \cdot \cos y \cdot y '}{1 + \sin y} ^ 2$

$y ' ' = \frac{5 {\pi}^{2} {e}^{\pi x} \cdot \left(1 + \sin y\right) - 5 \pi {e}^{\pi x} \cdot \cos y \frac{5 \pi {e}^{\pi x}}{1 + \sin y}}{1 + \sin y} ^ 2$

$y ' ' = \frac{5 {\pi}^{2} {e}^{\pi x} \cdot \left(1 + \sin y\right) - 5 \pi {e}^{\pi x} \cdot \cos y \frac{5 \pi {e}^{\pi x}}{1 + \sin y}}{1 + \sin y} ^ 2$

$y ' ' = \frac{5 {\pi}^{2} {e}^{\pi x} \cdot {\left(1 + \sin y\right)}^{2} - 25 {\pi}^{2} {e}^{2 \pi x} \cdot \cos y}{1 + \sin y} ^ 3$

$y ' ' = \frac{5 {\pi}^{2} {e}^{\pi x} \left({\left(1 + \sin y\right)}^{2} - 5 {e}^{\pi x} \cos y\right)}{1 + \sin y} ^ 3$