Quotient Rule
Let \(f\) and \(g\) be differentiable functions at the point \(a.\) If \(g(a) \neq 0,\) the quotient rule says that
\[\left(\frac{f}{g}\right)'(a) = \frac{f'(a)g(a)-f(a)g'(a)}{g^2(a)}\]
▼ Proof:
By the definition of derivative,
\begin{align}
\left(\frac{f}{g}\right)'(a) & = \lim_{h \rightarrow 0} \frac{\frac{f(a+h)}{g(a+h)}-\frac{f(a)}{g(a)}}{h} \\
& = \lim_{h \rightarrow 0} \frac{f(a+h)g(a)-f(a)g(a+h)}{hg(a)g(a+h)}
\end{align}
Now add and subtract \(f(a)g(a)\) from the numerator.
\begin{align}
\lim_{h \rightarrow 0} \frac{f(a+h)g(a)-f(a)g(a+h)}{hg(a)g(a+h)} & = \lim_{h \rightarrow 0} \frac{f(a+h)g(a)-f(a)g(a)+f(a)g(a)-f(a)g(a+h)}{hg(a)g(a+h)} \\
& = \lim_{h \rightarrow 0} \frac{(f(a+h)-f(a))g(a)+f(a)(g(a)-g(a+h))}{hg(a)g(a+h)} \\
& = \lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h} \cdot \frac{g(a)}{g(a)g(a+h)} - \frac{f(a)}{g(a)g(a+h)} \cdot \frac{g(a+h)-g(a)}{h} \\
& = f'(a) \cdot \frac{g(a)}{g(a)^2} - \frac{f(a)}{g(a)^2} \cdot g'(a) \\
& = \frac{f'(a)g(a)-f(a)g'(a)}{g(a)^2}
\end{align}
Examples
Find the derivative of \(\frac{x^2}{\text{sin}(x)}.\)
▼ Solution:
By the power rule, the derivative of \(x^2\) is \(2x.\) The derivative of \(\text{sin}(x)\) is \(\text{cos}(x).\) So, by the quotient rule,
\[\frac{d}{dx}\frac{x^2}{\text{sin}(x)} = \frac{2x\text{sin}(x) - x^2\text{cos}(x)}{\text{sin}(x)^2}\]
Find the derivative of \(\frac{e^x}{x^3}.\)
▼ Solution:
The derivative of \(e^x\) is \(e^x.\) By the power rule, the derivative of \(x^3\) is \(3x^2.\) So, by the quotient rule,
\[\frac{d}{dx}\frac{e^x}{x^3} = \frac{x^3e^x - 3x^2e^x}{x^6}\]
Find the derivative of \(\frac{\text{cos}(x)}{e^x}.\)
▼ Solution:
The derivative of \(\text{cos}(x)\) is \(-\text{sin}(x)\) and the derivative of \(e^x\) is \(e^x\). So, by the product rule,
\[\frac{d}{dx}\frac{\text{cos}(x)}{e^x} = \frac{-\text{sin}(x)e^x - \text{cos}(x)e^x}{e^{2x}}\]
This expression can be simplified to
\[-\frac{\text{sin}(x) + \text{cos}(x)}{e^{x}}\]