The product rule

Differentiation, derivatives and Taylor approximations: Rules for differentiating functions

The product rule

First have a look at one or more examples.

Given are the functions

$\begin{aligned}f(t) &= t^2+3\\ \\ g(t) &=t^2-4t+5\\ \\ p(t) &=f(t)\cdot g(t)=(t^2+3)\cdot(t^2-4t+5)\end{aligned}$ Calculate

$\begin{aligned}& f'(t)\\ \\ & g'(t)\\ \\ & p'(t)\quad \text{first expand brackets in }p(t))\\ \\ & f'(t)\cdot g(t)+f(t)\cdot g'(t)\end{aligned}$

Expansion of the brackets gives:

$p(t)=(t^2+3)\cdot(t^2-4t+5)=t^4-4 t^3+8 t^2-12 t+8$
So:

$\begin{aligned}f'(t) &= 2t\\ \\ g'(t) &= 2t-4\\ \\ p'(t) &=4 t^3-12 t^2+16 t-12\\ \\ f'(t)\cdot g(t)+f(t)\cdot g'(t) &= 2t\cdot (t^2-4t+5)+(t^2+3)\cdot ( 2t-4) \\ \\ &= (2t^3-8t^2+10t)+(2t^3-4t^2+6t-12) \\ \\ &=4 t^3-12 t^2+16 t-12 \end{aligned}$ Note that

$\left(f(t)\cdot g(t)\right)'=f'(t)\cdot g(t)+f(t)\cdot g'(t)$

New example

The pattern is clear and can be generalised to the product rule for differentiation.

Product rule

For functions $f$ and $g$ that are differentiabel at $x$ we have the product rule:

$\bigl(f(x)\cdot g(x)\bigr)'= f'(x)\cdot g(x)+f(x)\cdot g'(x)$

in short notation we have for differentiable functions

$\left(f\cdot g\right)'= f'\cdot g+f\cdot g'$

Example

$\begin{aligned}\\ &\bigl((x^2+2)(x^3+4)\bigr)'={}\\[0.25cm] &(x^2+2)'\cdot (x^3+4)+(x^2+2)\cdot (x^3+4)'={}\\[0.25cm] &2x\cdot (x^3+4)+(x^2+2)\cdot 3x^2={}\\[0.25cm] &5x^4+6x^2+8x\end{aligned}$

For math enthusiasts we present a proof of the product rule

$\left(f(x)\cdot g(x)\right)'= f'(x)\cdot g(x)+f(x)\cdot g'(x)$

For $p(x)=f(x)\cdot g(x)$ we have:

$\begin{aligned}\frac{{\vartriangle}p}{{\vartriangle}x} &= \frac{f(x+{\vartriangle}x)\cdot g(x+{\vartriangle}x)-f(x)\cdot g(x)}{{\vartriangle}x}\\ \\ &= \frac{f(x+{\vartriangle}x)\cdot g(x+{\vartriangle}x)-f(x)\cdot g(x+{\vartriangle}x)+f(x)\cdot g(x+{\vartriangle}x)-f(x)\cdot g(x)}{{\vartriangle}x}\\ \\ \\ &= \frac{f(x+{\vartriangle}x)\cdot g(x+{\vartriangle}x)-f(x)\cdot g(x+{\vartriangle}x)}{{\vartriangle}x}+\frac{f(x)\cdot g(x+{\vartriangle}x)-f(x)\cdot g(x)}{{\vartriangle}x}\\ \\ &= \frac{{\vartriangle}f}{{\vartriangle}x}\cdot g(x+{\vartriangle}x)+f(x)\cdot\frac{{\vartriangle}g}{{\vartriangle}x}\\ \\ &\to f'(x)\cdot g(x)+ f(x)\cdot g'(x)\quad\text{when }{{\vartriangle}x}\to 0\end{aligned}$

You can also write the product rule in a homogeneous way:

$\frac{(f\cdot g)'}{f\cdot g}= \frac{f'}{f}+\frac{g'}{g}$ In this form it is easy to see what should be done for a product of three functions:

$\frac{\left(f\cdot g\cdot h\right)'}{f\cdot g\cdot h}= \frac{f'}{f}+\frac{g'}{g}+ \frac{h'}{h}$

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

The above example of the product rule conveys the strategy of calculating the derivative of a product of two differentiable functions $f\cdot g$ :

Determine how to split the given function as a product of two functions $f\cdot g$ , that is, distinguish $f$ and $g$ ;
Calculate the derivatives $f'$ and $g'$ ;
Apply the product rule $\left(f\cdot g\right)'= f'\cdot g+f\cdot g'$ ;
Work out and/or simplify the calculated derivative.

Mathcentre video

Product Rule (14:37)