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 \[\begin{aligned}f(t) &= t^2-2\\ \\ g(t) &=t^2+2t+3\\ \\ p(t) &=f(t)\cdot g(t)=(t^2-2)\cdot(t^2+2t+3)\end{aligned}\] 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: \[p(t)=(t^2-2)\cdot(t^2+2t+3)=t^4+2 t^3+t^2-4 t+1\]
So: \[\begin{aligned}f'(t) &= 2t\\ \\
g'(t) &= 2t+2\\ \\
p'(t) &=4 t^3+6 t^2+2 t-4\\ \\
f'(t)\cdot g(t)+f(t)\cdot g'(t) &= 2t\cdot (t^2+2t+3)+(t^2-2)\cdot ( 2t+2) \\ \\
&= (2t^3+4t^2+6t)+(2t^3+2t^2-4t-4) \\ \\
&=4 t^3+6 t^2+2 t-4
\end{aligned}\] 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)