The product rule for differentiation

Rules of differentiation: Rules of computation for the derivative

The product rule for differentiation

Product of two functions

Let $f$ and $g$ be two functions.

The product of $f$ and $g$ is the function that assigns to $x$ the value $f(x)\cdot g(x)$ . This function is denoted by $f\cdot g$ , so the function rule is $(f\cdot g)(x) = f(x)\cdot g(x)$ .

For example, if $f(x)=x+1$ and $g(x)=x^3+1$ , then $f\cdot g$ is the function with function rule $\left(f\cdot g\right)(x)=f(x)\cdot g(x) = \left(x+1\right)\cdot \left(x^3+1\right) = x^4+x^3+x+1\tiny.$

Product rule for differentiation

The derivative $(f\cdot g)'$ of the product $f\cdot g$ is given by the productformula

$(f\cdot g)' = f' \cdot g + f\cdot g'\tiny.$

This means that $(f\cdot g)'(x) = f'(x)\cdot g(x) + f(x)\cdot g'(x)$ for all $x$ .

In order to prove the rule, we rewrite difference quotient as follows:

$\begin{array}{rcl} \text{difference quotient}&=&\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\ &=& \frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h}\\ &=& \frac{(f(x+h)-f(x))g(x)+f(x+h)(g(x+h)-g(x))}{h}\\ &=& \frac{f(x+h)-f(x)}{h}g(x)+f(x+h)\frac{g(x+h)-g(x)}{h}\\ \end{array}$

Now we can calculate the limit of the difference quotient for $h\to 0$ : $\begin{array}{rcl} \left(f\cdot g\right)'(x) &=& \lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\ &=&\lim_{h\to 0}\left( \frac{f(x+h)-f(x)}{h}g(x)+f(x+h)\frac{g(x+h)-g(x)}{h}\right)\\ &=&\lim_{h\to 0}\left(\frac{f(x+h)-f(x)}{h}g(x)\right)+\lim_{h\to 0}\left(f(x+h)\frac{g(x+h)-g(x)}{h}\right)\\ &=&\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\cdot \lim_{h\to 0} g(x)+\lim_{h\to 0}f(x+h)\cdot \lim_{h\to 0}\frac{g(x+h)-g(x)}{h}\\ &=& f'(x)\cdot g(x)+f(x)\cdot g'(x)\tiny.\\ \end{array}$

Determine the derivative of the function $(x^2+4\cdot x+4)\cdot( x^3+4\cdot x-2)$ .

$5\cdot x^4+16\cdot x^3+24\cdot x^2+28\cdot x+8$

The product rule gives that the derivative is equal to $\frac{\dd}{\dd x}(x^2+4\cdot x+4)\cdot (x^3+4\cdot x-2) + (x^2+4\cdot x+4)\cdot\frac{\dd}{\dd x} (x^3+4\cdot x-2)\tiny.$ According to The derivative of a polynomial this is equal to
$(2x+4)\cdot (x^3+4\cdot x-2) + (x^2+4\cdot x+4)\cdot (3x^2+4)$ , which can be rewritten as $2\cdot x^4+4\cdot x^3+8\cdot x^2+12\cdot x-8+3\cdot x^4+12\cdot x^3+16\cdot x^2+16\cdot x+16$
and thus is equal to $5\cdot x^4+16\cdot x^3+24\cdot x^2+28\cdot x+8$ .

It is also possible to start by calculating the product of $x^2+4\cdot x+4$ and $x^3+4\cdot x-2$ ; this is $(x^2+4\cdot x+4)\cdot(x^3+4\cdot x-2) = x^5+4\cdot x^4+8\cdot x^3+14\cdot x^2+8\cdot x-8$ . The derivative then follows from The derivative of a polynomial.

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