 # How to Find Derivatives using the Product Rule

The product rule is the rule you use when you have a product of two or more functions.

Let $u\left(x\right)$ and $v\left(x\right)$ be two functions of $x$. The product rule tells you how to differentiate the product of these two functions. The product is the new function $u\left(x\right)\cdot v\left(x\right)$. To simplify it, you just write $u$ and $v$ for the functions $u\left(x\right)$ and $v\left(x\right)$, but it’s important to remember that these are functions of $x$.

Formula

### TheProductRule

 ${\left(uv\right)}^{\prime }={u}^{\prime }v+u{v}^{\prime },$

where $u=u\left(x\right)$ and $v=v\left(x\right)$.

Example 1

Differentiate the expression ${x}^{2}\sqrt{2x}$

Looking at the expression, you decide that $u={x}^{2}$ and $v=\sqrt{2x}$. That gives you ${u}^{\prime }=2x$ and ${v}^{\prime }=\frac{1}{\sqrt{2x}}$, and the differentiation is $\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}{\left({x}^{2}\sqrt{2x}\right)}^{\prime }& =\phantom{\rule{-0.17em}{0ex}}{\left({x}^{2}\right)}^{\prime }\cdot \sqrt{2x}+{x}^{2}\cdot \phantom{\rule{-0.17em}{0ex}}{\left(\sqrt{2x}\right)}^{\prime }\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2x\cdot \sqrt{2x}+{x}^{2}\cdot \frac{1}{\sqrt{2x}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2x\sqrt{2x}+\frac{{x}^{2}}{\sqrt{2x}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{2x\sqrt{2x}\cdot \sqrt{2x}}{\sqrt{2x}}+\frac{{x}^{2}}{\sqrt{2x}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{4{x}^{2}}{\sqrt{2x}}+\frac{{x}^{2}}{\sqrt{2x}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{5{x}^{2}}{\sqrt{2x}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 2

Differentiate the expression ${e}^{x}\phantom{\rule{-0.17em}{0ex}}\left(2{x}^{3}+3x\right)$

Here, you decide to say that $u={e}^{x}$ and $v=2{x}^{3}+3x$. That gives you ${u}^{\prime }={e}^{x}$ and ${v}^{\prime }=6{x}^{2}+3$, and the differentiation becomes

$\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}{\left[{e}^{x}\phantom{\rule{-0.17em}{0ex}}\left(2{x}^{3}+3x\right)\right]}^{\prime }& =\phantom{\rule{-0.17em}{0ex}}{\left({e}^{x}\right)}^{\prime }\cdot \phantom{\rule{-0.17em}{0ex}}\left(2{x}^{3}+3x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}+{e}^{x}\cdot \phantom{\rule{-0.17em}{0ex}}{\left(2{x}^{3}+3x\right)}^{\prime }\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={e}^{x}\cdot \phantom{\rule{-0.17em}{0ex}}\left(2{x}^{3}+3x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}+{e}^{x}\cdot \phantom{\rule{-0.17em}{0ex}}\left(6{x}^{2}+3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2{x}^{3}{e}^{x}+3x{e}^{x}+6{x}^{2}{e}^{x}+3{e}^{x}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={e}^{x}\phantom{\rule{-0.17em}{0ex}}\left(2{x}^{3}+6{x}^{2}+3x+3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}{\left[{e}^{x}\phantom{\rule{-0.17em}{0ex}}\left(2{x}^{3}+3x\right)\right]}^{\prime }& =\phantom{\rule{-0.17em}{0ex}}{\left({e}^{x}\right)}^{\prime }\cdot \phantom{\rule{-0.17em}{0ex}}\left(2{x}^{3}+3x\right)+{e}^{x}\cdot \phantom{\rule{-0.17em}{0ex}}{\left(2{x}^{3}+3x\right)}^{\prime }\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={e}^{x}\cdot \phantom{\rule{-0.17em}{0ex}}\left(2{x}^{3}+3x\right)+{e}^{x}\cdot \phantom{\rule{-0.17em}{0ex}}\left(6{x}^{2}+3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2{x}^{3}{e}^{x}+3x{e}^{x}+6{x}^{2}{e}^{x}+3{e}^{x}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={e}^{x}\phantom{\rule{-0.17em}{0ex}}\left(2{x}^{3}+6{x}^{2}+3x+3\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 3

Differentiate the expression $\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}+x-1\right)\mathrm{ln}x$

Here, you decide that $u={x}^{2}+x-1$ and $v=\mathrm{ln}x$. That gives you ${u}^{\prime }=2x+1$ and ${v}^{\prime }=\frac{1}{x}$, and the differentiation becomes

$\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}{\left[\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}+x-1\right)\mathrm{ln}x\right]}^{\prime }& =\phantom{\rule{-0.17em}{0ex}}{\left({x}^{2}+x-1\right)}^{\prime }\cdot \mathrm{ln}x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{=}+\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}+x-1\right)\cdot {\left(\mathrm{ln}x\right)}^{\prime }\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(2x+1\right)\cdot \mathrm{ln}x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{=}+\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}+x-1\right)\cdot \frac{1}{x}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2x\mathrm{ln}x+\mathrm{ln}x+x+1-\frac{1}{x}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}{\left[\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}+x-1\right)\mathrm{ln}x\right]}^{\prime }& =\phantom{\rule{-0.17em}{0ex}}{\left({x}^{2}+x-1\right)}^{\prime }\cdot \mathrm{ln}x+\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}+x-1\right)\cdot {\left(\mathrm{ln}x\right)}^{\prime }\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(2x+1\right)\cdot \mathrm{ln}x+\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}+x-1\right)\cdot \frac{1}{x}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2x\mathrm{ln}x+\mathrm{ln}x+x+1-\frac{1}{x}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$