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
\[\left(\frac{1}{x^2+1}\right)'=-\frac{2x}{(x^2+1)^2}\]
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
\[\begin{aligned}\left(\frac{3x^2+2}{x^2+1}\right)' &=\frac{6x\cdot (x^2+1)-(3x^2+2)\cdot 2x}{(x^2+1)^2}\\[0.25cm] &=\frac{2x}{(x^2+1)^2}\\[0.5cm]
\left(\frac{3x-4}{2x+1}\right)' &=\frac{3\cdot (2x+1)-(3x-4)\cdot 2}{(x^2+1)^2}\\[0.25cm] &=\frac{11x}{(x+1)^2}\quad\text{provided }x\neq -1\end{aligned}\]
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.
g'(x) &= 3\cdot 2x= 6x\\ \\
q'(x) &=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{g(x)^2} \\ \\
&= \frac{-4\cdot(3x^2+1)+4x\cdot 6x}{(3x^2+1)^2}\\ \\
&= \frac{12x^2-4}{(3x^2+1)^2} \end{aligned}\]
Mathcentre video
Quotient Rule (16:47)
