The quotient rule

Differentiation, derivatives and Taylor approximations: Rules for differentiating functions

The quotient rule

We start looking at a special case of the quotient rule of differentiation:

Reciprocal rule

If \(g\) is differentiable at \(x\) and \(g(x)\neq 0\), then the reciprocal function \(q=1/g\) is also differentiable at \(x\), and \[q'(x)= -\frac{g'(x)}{g(x)^2}\]

Example

\[\left(\frac{1}{x^2+1}\right)'=-\frac{2x}{(x^2+1)^2}\]

For math enthusiasts we present a proof: \[\begin{aligned}\frac{{\vartriangle}q}{{\vartriangle}x} &= \frac{q(x+{\vartriangle}x)-q(x)}{{\vartriangle}x} \\ \\
&=\left(\frac{1}{g(x+{\vartriangle}x)}-\frac{1}{g(x)}\right)\cdot \frac{1}{{\vartriangle}x} \\ \\
&=\left(\frac{g(x)-g(x+{\vartriangle}x)}{g(x+{\vartriangle}x)\cdot g(x)}\right)\cdot \frac{1}{{\vartriangle}x}\\ \\
&=-\frac{g(x+{\vartriangle}x)-g(x)}{{\vartriangle}x}\cdot \frac{1}{g(x+{\vartriangle}x)\cdot g(x)}\\ \\
&=-\frac{{\vartriangle}g}{{\vartriangle}x}\cdot \frac{1}{g(x+{\vartriangle}x)\cdot g(x)}\\ \\
&\to -\frac{g'(x)}{g(x)^2}\quad\mathrm{als\;}{\vartriangle}x\to 0\end{aligned}\]

The general quotient rule for differentiation:

Quotient Rule

If \(f\) and \(g\) are functions differentiable at \(x\), and if \(g(x)\neq 0\), then the quotient \(f/g\) is also differentiable at \(x\), and \[\left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{g(x)^2}\]

In short notation for differentiable functions: \(\displaystyle \left(\frac{f}{g}\right)'=\frac{f'\cdot g-f\cdot g'}{g^2}\).

Examples

\[\begin{aligned}\left(\frac{3x^2+2}{x^2+1}\right)' &=\frac{6x\cdot (x^2+1)-(3x^2+2)\cdot 2x}{(x^2+1)^2}\\[0.25cm] &=\frac{2x}{(x^2+1)^2}\\[0.5cm]
\left(\frac{3x-4}{2x+1}\right)' &=\frac{3\cdot (2x+1)-(3x-4)\cdot 2}{(x^2+1)^2}\\[0.25cm] &=\frac{11x}{(x+1)^2}\quad\text{provided }x\neq -1\end{aligned}\]

For math enthusiasts we present a proof of the quotient rule \[\left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{g(x)^2}\]

For \(\displaystyle q(x)=\frac{f(x)}{g(x)}\) we get from the product rule and the above special case: \[\begin{aligned}q'(x)
&= \left(f(x)\cdot \frac{1}{g(x)}\right)'\\ \\
&= f'(x)\cdot \frac{1}{g(x)}+ f(x)\cdot \left(\frac{1}{g(x)}\right)' \\ \\
&= \frac{f'(x)}{g(x)}+ f(x)\cdot \left(-\frac{g'(x)}{g(x)^2}\right) \\ \\
&= \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{g(x)^2}
\end{aligned}\]

Mnemonic Mind theminus sign in the numerator of the derivative of a quotient of two differentiable function. A mnemonic for remembering the quotient rule is the following yodeling-type chant "low dee high minus high dee low." For the whole formula of the derivative you may say to yourself "low dee high minus high dee low over the square of what's below

Below are a few examples that illustrate the quotient rule. They all follow the same strategy.

How to calculate the derivative of a quotient of two differentiable functions

A strategy for calculating the derivative of a product of two differentiable functions\(f/g\) could be:

Determine how the given function can be understood as a quotient \(f/g\), that is, distinguish \(f\) and \(g\);
Calculate the derivatives \(f'\) and \(g'\);
Apply the quotient rule \(\displaystyle\left(\frac{f}{g}\right)'=\frac{f'\cdot g-f\cdot g'}{g^2}\);
Work out and/or simplify the calculated derivative.

Use the quotient rule to calculate the derivative of \[q(x)=\frac{f(x)}{g(x)}\] where \[f(x)=4x,\quad g(x)=5x^2+1\] Do this as follows: first calculate \(f'(x)\) and \(g'(x)\),
then calculate the derivative of the quotient, i.e. \(q'(x)\), and finally simplify the result..

\[\begin{aligned} f'(x)&=4 \\ \\
g'(x) &= 5\cdot 2x= 10x\\ \\
q'(x) &=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{g(x)^2} \\ \\
&= \frac{4\cdot(5x^2+1)-4x\cdot 10x}{(5x^2+1)^2}\\ \\
&= \frac{-20x^2+4}{(5x^2+1)^2} \end{aligned}\]

New example

Mathcentre video

Quotient Rule (16:47)