# Product rule

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

${\displaystyle F(x)=f(x)g(x)}$

${\displaystyle F'(x)=f'(x)g(x)+g'(x)f(x)}$.

## Steps

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

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

We can set

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

${\displaystyle g(x)=\cos(x)}$.

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

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

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

Now we can substitute the values into the equation,

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

## Proof

One definition of a derivative is

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

${\displaystyle \lim _{h\to 0}{\frac {f(x+h)g(x+h)-f(x)g(x)}{h}}}$

We can't really do much with this so we need to manipulate the equation.

${\displaystyle \lim _{h\to 0}{\frac {f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h}}}$

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

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

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

${\displaystyle F'(x)=f'(x)g(x)+g'(x)f(x)}$.

## References

1. "Product rule proof (video) | Optional videos". Khan Academy. Retrieved 2022-09-12.