# Product rule

In differential calculus, the product rule is a rule that helps calculate derivates that have multiplication.

$F(x)=f(x)g(x)$ $F'(x)=f'(x)g(x)+g'(x)f(x)$ .

## Steps

Say we have the function $F(x)=(x^{2}+3)\cos(x)$ .

The two functions being multiplied are $x^{2}+3$ and $\cos(x)$ .

We can set

$f(x)=x^{2}+3$ and

$g(x)=\cos(x)$ .

The rule needs us to find the derivative of both $f(x)$ and $g(x)$ .

We can find $f'(x)$ by first using the sum rule to split $f(x)$ into $x^{2}$ and $3$ . After using the power rule, we have $f'(x)=2x$ .

To find $g'(x)$ , we need to find the derivative of $\cos(x)$ , which is $-\sin(x)$ , meaning $g'(x)=-\sin(x)$ .

Now we can substitute the values into the equation,

$F'(x)=2x\cos(x)-\sin(x)(x^{2}+3)$ .

## Proof

One definition of a derivative is

$F'(x)=\lim _{h\to 0}{\frac {F(x+h)-F(x)}{h}}$ , and we're trying to find the derivative of $f(x)g(x)$ , so we can first set $F(x)$ to $f(x)g(x)$ .

$\lim _{h\to 0}{\frac {f(x+h)g(x+h)-f(x)g(x)}{h}}$ We can't really do much with this so we need to manipulate the equation.

$\lim _{h\to 0}{\frac {f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h}}$ The $-f(x+h)g(x)+f(x+h)g(x)$ part is equal to $0$ , meaning it didn't change the value of the equation. Now we can factor,

$\lim _{h\to 0}{f(x+h)}\lim _{h\to 0}{\frac {g(x+h)-g(x)}{h}}+\lim _{h\to 0}{g(x)}\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$ , and because $h$ approaches $0$ , $\lim _{h\to 0}{f(x+h)}$ is equal to $f(x)$ .

${f(x+h)}\lim _{h\to 0}{\frac {g(x+h)-g(x)}{h}}+{g(x)}\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$ , and $\lim _{h\to 0}{\frac {g(x+h)-g(x)}{h}}$ and $\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$ are just equal to $g'(x)$ and $f'(x)$ .

$F'(x)=f'(x)g(x)+g'(x)f(x)$ .