# Quotient Rule

## Related calculator: Online Derivative Calculator with Steps

Quotient Rule. If $f$ and $g$ are differentiable, it can be stated that ${\left(\frac{{{f{{\left({x}\right)}}}}}{{{g{{\left({x}\right)}}}}}\right)}'=\frac{{{f{'}}{\left({x}\right)}{g{{\left({x}\right)}}}-{f{{\left({x}\right)}}}{g{'}}{\left({x}\right)}}}{{{\left({g{{\left({x}\right)}}}\right)}}^{{2}}}$.

Proof.

Let ${h}{\left({x}\right)}=\frac{{{f{{\left({x}\right)}}}}}{{{g{{\left({x}\right)}}}}}$; then, ${f{{\left({x}\right)}}}={h}{\left({x}\right)}{g{{\left({x}\right)}}}$.

Differentiating both sides gives: ${f{'}}{\left({x}\right)}={\left({h}{\left({x}\right)}{g{{\left({x}\right)}}}\right)}'$, or, using the product rule, ${f{'}}{\left({x}\right)}={h}{\left({x}\right)}{g{'}}{\left({x}\right)}+{h}'{\left({x}\right)}{g{{\left({x}\right)}}}$.

From this, we have that ${h}'{\left({x}\right)}=\frac{{{f{'}}{\left({x}\right)}-{h}{\left({x}\right)}{g{'}}{\left({x}\right)}}}{{{g{{\left({x}\right)}}}}}$.

Remembering that ${h}{\left({x}\right)}=\frac{{{f{{\left({x}\right)}}}}}{{{g{{\left({x}\right)}}}}}$, we can rewrite the last identity as ${\left(\frac{{{f{{\left({x}\right)}}}}}{{{g{{\left({x}\right)}}}}}\right)}'=\frac{{{f{'}}{\left({x}\right)}-\frac{{{f{{\left({x}\right)}}}}}{{{g{{\left({x}\right)}}}}}{g{'}}{\left({x}\right)}}}{{{g{{\left({x}\right)}}}}}$, or ${\left(\frac{{{f{{\left({x}\right)}}}}}{{{g{{\left({x}\right)}}}}}\right)}'=\frac{{{f{'}}{\left({x}\right)}{g{{\left({x}\right)}}}-{f{{\left({x}\right)}}}{g{'}}{\left({x}\right)}}}{{{\left({g{{\left({x}\right)}}}\right)}}^{{2}}}$.

Example 1. Find the derivative of ${f{{\left({x}\right)}}}=\frac{{{{e}}^{{x}}}}{{{{x}}^{{3}}+{x}}}$.

${f{'}}{\left({x}\right)}={\left(\frac{{{{e}}^{{x}}}}{{{{x}}^{{3}}+{x}}}\right)}'=\frac{{{\left({{e}}^{{x}}\right)}'{\left({{x}}^{{3}}+{x}\right)}-{{e}}^{{x}}{\left({{x}}^{{3}}+{x}\right)}'}}{{{\left({{x}}^{{3}}+{x}\right)}}^{{2}}}=\frac{{{{e}}^{{x}}{\left({{x}}^{{3}}+{x}\right)}-{{e}}^{{x}}{\left({3}{{x}}^{{2}}+{1}\right)}}}{{{\left({{x}}^{{3}}+{x}\right)}}^{{2}}}=$

$=\frac{{{{e}}^{{x}}{\left({{x}}^{{3}}-{3}{{x}}^{{2}}+{x}-{1}\right)}}}{{{\left({{x}}^{{3}}+{x}\right)}}^{{2}}}$.

Let's do another example.

Example 2. Find the derivative of ${f{{\left({t}\right)}}}=\frac{{{{t}}^{{5}}+{1}}}{{{t}}^{{3}}}$.

${f{'}}{\left({t}\right)}={\left(\frac{{{{t}}^{{5}}+{1}}}{{{{t}}^{{3}}}}\right)}'=\frac{{{\left({{t}}^{{5}}+{1}\right)}'{{t}}^{{3}}-{\left({{t}}^{{5}}+{1}\right)}{\left({{t}}^{{3}}\right)}'}}{{{\left({{t}}^{{3}}\right)}}^{{2}}}=\frac{{{5}{{t}}^{{4}}{{t}}^{{3}}-{\left({{t}}^{{5}}+{1}\right)}\cdot{3}{{t}}^{{2}}}}{{{t}}^{{6}}}=$

$=\frac{{{5}{{t}}^{{7}}-{3}{{t}}^{{7}}-{3}{{t}}^{{2}}}}{{{{t}}^{{6}}}}=\frac{{{2}{{t}}^{{7}}-{3}{{t}}^{{2}}}}{{{{t}}^{{6}}}}={2}{t}-\frac{{3}}{{{t}}^{{4}}}$.

Note, however, that Example 2 can be used without the quotient rule.

Since $\frac{{{{t}}^{{5}}+{1}}}{{{t}}^{{3}}}={{t}}^{{2}}+\frac{{1}}{{{t}}^{{3}}}={{t}}^{{2}}+{{t}}^{{-{3}}}$, we have that ${f{'}}{\left({t}\right)}={2}{{t}}^{{{2}-{1}}}-{3}{{t}}^{{-{3}-{1}}}={2}{t}-{3}{{t}}^{{-{4}}}={2}{t}-\frac{{3}}{{{t}}^{{4}}}$.

Example 2 showed that first you need to look at the function and make sure that the quotient rule is the only way to find the derivative (like in Example 1).

The general advice is not to use the quotient rule every time you see a quotient. It is a rather hard technique, and if there are other ways to find the derivative, use them.