How to Differentiate Functions Using the Quotient Rule

The quotient rule is the rule that tells you how to take the derivative of a function that is a ratio of two differentiable functions.

Formula

TheQuotientRule

 $\phantom{\rule{-0.17em}{0ex}}{\left(\frac{u}{v}\right)}^{\prime }=\frac{{u}^{\prime }v-u{v}^{\prime }}{{v}^{2}}$

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

Note! Sometimes you get lucky, and it’s possible to simplify the answer. Most of the time that’s not possible though, and you can just leave the fraction as it is.

Example 1

Differentiate the expression $\frac{2x+1}{{e}^{x}}$

Here, $u=2x+1$ and $v={e}^{x}$. You get ${u}^{\prime }=2$ and ${v}^{\prime }={e}^{x}$, which gives you

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

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

Example 2

Differentiate the expression $\frac{3{x}^{3}-2{x}^{2}+7}{x-1}$

Here, you have $u=3{x}^{3}-2{x}^{2}+7$ and $v=x-1$. That means ${u}^{\prime }=9{x}^{2}-2$ and ${v}^{\prime }=1$, and the derivative is

$\begin{array}{llll}\hfill & \phantom{=}\phantom{\rule{-0.17em}{0ex}}{\left(\frac{3{x}^{3}-2{x}^{2}+7}{x-1}\right)}^{\prime }\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{{\left(x-1\right)}^{2}}\left(\phantom{\rule{-0.17em}{0ex}}{\left(3{x}^{3}-2{x}^{2}+7\right)}^{\prime }\cdot \left(x-1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}-\phantom{\rule{-0.17em}{0ex}}\left(3{x}^{3}-2{x}^{2}+7\right){\left(x-1\right)}^{\prime }\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\phantom{\rule{-0.17em}{0ex}}\left(9{x}^{2}-4x\right)\cdot \left(x-1\right)-\phantom{\rule{-0.17em}{0ex}}\left(3{x}^{3}-2{x}^{2}+7\right)\cdot 1}{{\left(x-1\right)}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{9{x}^{3}-9{x}^{2}-4{x}^{2}+4x-3{x}^{3}+2{x}^{2}-7}{{\left(x-1\right)}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{6{x}^{3}-11{x}^{2}+4x-7}{{\left(x-1\right)}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}{\left(\frac{3{x}^{3}-2{x}^{2}+7}{x-1}\right)}^{\prime }& =\frac{\phantom{\rule{-0.17em}{0ex}}{\left(3{x}^{3}-2{x}^{2}+7\right)}^{\prime }\cdot \left(x-1\right)-\phantom{\rule{-0.17em}{0ex}}\left(3{x}^{3}-2{x}^{2}+7\right){\left(x-1\right)}^{\prime }}{{\left(x-1\right)}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\phantom{\rule{-0.17em}{0ex}}\left(9{x}^{2}-4x\right)\cdot \left(x-1\right)-\phantom{\rule{-0.17em}{0ex}}\left(3{x}^{3}-2{x}^{2}+7\right)\cdot 1}{{\left(x-1\right)}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{9{x}^{3}-9{x}^{2}-4{x}^{2}+4x-3{x}^{3}+2{x}^{2}-7}{{\left(x-1\right)}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{6{x}^{3}-11{x}^{2}+4x-7}{{\left(x-1\right)}^{2}}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

As $x=1$ is not a root of the numerator, you can’t simplify the expression.

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