# How do you differentiate f(x)=e^x*sin(5x^2) using the product rule?

Apr 28, 2018

${e}^{x} \left(\sin \left(5 {x}^{2}\right) + 10 x \cos \left(5 {x}^{2}\right)\right)$

#### Explanation:

Let $u = {e}^{x}$ and $v = \sin \left(5 {x}^{2}\right)$
Therefore $\frac{\mathrm{du}}{\mathrm{dx}} = {e}^{x}$
and to find $\frac{\mathrm{dv}}{\mathrm{dx}}$ you need to use chain rule...
so... let $w = 5 {x}^{2}$ then $v = \sin \left(w\right)$
$\frac{\mathrm{dv}}{\mathrm{dw}} = \cos \left(w\right)$
and $\frac{\mathrm{dw}}{\mathrm{dx}} = 10 x$
so the differential of $v = \sin \left(5 {x}^{2}\right)$ is $10 x \cos \left(5 {x}^{2}\right)$

from there you use the rest of product rule so... $\frac{\mathrm{dy}}{\mathrm{dx}} = \left(v \cdot \frac{\mathrm{du}}{\mathrm{dx}}\right) + \left(u \cdot \frac{\mathrm{dv}}{\mathrm{dx}}\right)$
so...
$\left(\sin \left(5 {x}^{2}\right) \cdot {e}^{x}\right) + \left({e}^{x} \cdot 10 x \cos \left(5 {x}^{2}\right)\right)$
${e}^{x} \left(\sin \left(5 {x}^{2}\right) + 10 x \cos \left(5 {x}^{2}\right)\right)$

Apr 28, 2018

$f ' \left(x\right) = {e}^{x} \left(10 x \cos \left(5 {x}^{2}\right) + \sin \left(5 {x}^{2}\right)\right)$

#### Explanation:

$\text{Given "f(x)=g(x)h(x)" then}$

$f ' \left(x\right) = g \left(x\right) h ' \left(x\right) + h \left(x\right) g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{product rule}}$

$g \left(x\right) = {e}^{x} \Rightarrow g ' \left(x\right) = {e}^{x}$

$h \left(x\right) = \sin \left(5 {x}^{2}\right) \leftarrow \textcolor{b l u e}{\text{use chain rule}}$

$\Rightarrow h ' \left(x\right) = \cos \left(5 {x}^{2}\right) \times \frac{d}{\mathrm{dx}} \left(5 {x}^{2}\right)$

$\textcolor{w h i t e}{\Rightarrow h ' \left(x\right)} = 10 x \cos \left(5 {x}^{2}\right)$

$\Rightarrow f ' \left(x\right) = 10 x \cos \left(5 {x}^{2}\right) {e}^{x} + \sin \left(5 {x}^{2}\right) {e}^{x}$

$\textcolor{w h i t e}{\Rightarrow f ' \left(x\right)} = {e}^{x} \left(10 x \cos \left(5 {x}^{2}\right) + \sin \left(5 {x}^{2}\right)\right)$