Quotient Rule Differentiation

The quotient rule is a formula for taking the derivative of a quotient of two functions. This formula makes it somewhat easier to keep track of all of the terms.
If $u(x)$ and $v(x)$ are two functions of $x$ and $\displaystyle f(x)=\dfrac{u(x)}{v(x)}$, then
$$ \large f^{\prime}(x)=\displaystyle \dfrac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v(x)^2} $$
Expressions like $\displaystyle \dfrac{x^2+x+1}{4x-2}$, $\displaystyle \dfrac{\sqrt{x+2}}{x^2-4}$ and $\displaystyle \dfrac{x^4}{(x^3-x^2-1)^5}$ are called quotients because they represent the division of one function by another.

Some students prefer to apply the product rule instead of the quotient rule. For instance, $\displaystyle \dfrac{x}{x+1}$ can be changed to $x(x+1)^{-1}$, but this is not a great idea to differentiate as many students made careless mistakes while applying product rule due to taking longer steps. A form of quotient must apply the quotient rule properly if you want to reduce silly mistakes!

Example 1

Use the quotient rule to differentiate $\displaystyle f(x)=\dfrac{x}{x+1}$.

$ \begin{align} \displaystyle
f^{\prime}(x) &= \dfrac{x^{\prime} \times (x+1)-x \times (x+1)^{\prime}}{(x+1)^2} \\
&= \dfrac{1 \times (x+1)-x \times 1}{(x+1)^2} \\
&= \dfrac{(x+1)-x}{(x+1)^2} \\
&= \dfrac{1}{(x+1)^2}
\end{align} $

Example 2

Use the quotient rule to differentiate $\displaystyle f(x)=\dfrac{3x+1}{1-x}$.

$ \begin{align} \displaystyle
f^{\prime}(x) &= \dfrac{(3x+1)^{\prime} \times (1-x)-(3x+1) \times (1-x)^{\prime}}{(1-x)^2} \\
&= \dfrac{3 \times (1-x)-(3x+1) \times (-1)}{(1-x)^2} \\
&= \dfrac{3-3x +3x+1}{(1-x)^2} \\
&= \dfrac{4}{(1-x)^2}
\end{align} $

Example 3

Use the quotient rule to differentiate $\displaystyle f(x)=\dfrac{3x+1}{1-x}$.

Example 4

Use the quotient rule to differentiate $\displaystyle f(x)=\dfrac{(2x+1)^3}{(4x-1)^4}$.

$ \begin{align} \displaystyle
f^{\prime}(x) &= \dfrac{\big[(2x+1)^3\big]^{\prime} \times (4x-1)^4-(2x+1)^3 \times \big[(4x-1)^4\big]^{\prime}}{\big[(4x-1)^4\big]^2} \\
&= \dfrac{3(2x+1)^{3-1} \times (2x+1)^{\prime} \times (4x-1)^4-(2x+1)^3 \times 4(4x-1)^{4-1} \times (4x-1)^{\prime}}{(4x-1)^8} \\
&= \dfrac{3(2x+1)^2 \times 2 \times (4x-1)^4-(2x+1)^3 \times 4(4x-1)^3 \times 4}{(4x-1)^8} \\
&= \dfrac{6(2x+1)^2 (4x-1)^4-16(2x+1)^3 (4x-1)^3}{(4x-1)^8} \\
&= \dfrac{(2x+1)^2(4x-1)^3\big[6(4x-1)-16(2x+1)\big]}{(4x-1)^8} \\
&= \dfrac{(2x+1)^2(4x-1)^3(24x-6-32x-16)}{(4x-1)^8} \\
&= \dfrac{(2x+1)^2(4x-1)^3(-8x-22)}{(4x-1)^8} \\
&= \dfrac{(2x+1)^2(-8x-22)}{(4x-1)^5}
\end{align} $

Extension Examples

These Extension Examples require to have some prerequisite skills, including;
$ \begin{align} \displaystyle
\dfrac{d}{dx}\sin{x} &= \cos{x} \\
\dfrac{d}{dx}\cos{x} &= -\sin{x} \\
\dfrac{d}{dx}e^x &= e^x \\
\dfrac{d}{dx}\log_e{x} &= \dfrac{1}{x} \\
\end{align} $

Example 5

Find $\displaystyle \dfrac{dy}{dx}$ of $\displaystyle y=\dfrac{\sin{x}}{\cos{x}}$, known that $\dfrac{d}{dx}\sin{x} = \cos{x}$ and $\dfrac{d}{dx}\cos{x} = -\sin{x}$.

$ \begin{align} \displaystyle
\dfrac{dy}{dx} &= \dfrac{\dfrac{d}{dx}\sin{x} \times \cos{x}-\sin{x} \times \dfrac{d}{dx}\cos{x}}{\cos^2{x}} \\
&= \dfrac{\cos{x} \times \cos{x}-\sin{x} \times (-\sin{x})}{\cos^2{x}} \\
&= \dfrac{\cos^2{x} + \sin^2{x}}{\cos^2{x}}
\end{align} $

Note that optionally $\cos^2{x} + \sin^2{x}$ can be simplified further to $1$, as $\cos^2{x} + \sin^2{x}=1$ in one of the trigonometric properties, however we do not cover this skill here.

Example 6

Find $\displaystyle \dfrac{dy}{dx}$ of $\displaystyle y=\dfrac{e^x}{\log_e{x}}$, known that $\dfrac{d}{dx}e^x = e^x$ and $\dfrac{d}{dx}\log_e{x} = \dfrac{1}{x}$.

$ \begin{align} \displaystyle
\dfrac{dy}{dx} &= \dfrac{\dfrac{d}{dx}e^x \times \log_e{x}-e^x \times \dfrac{d}{dx}\log_e{x}}{(\log_e{x})^2} \\
&= \dfrac{e^x \times \log_e{x}-e^x \times \dfrac{1}{x}}{(\log_e{x})^2} \\
&= \dfrac{xe^x\log_e{x}-e^x}{x(\log_e{x})^2}
\end{align} $

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