Product rule

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

$F(x)=f(x)g(x)$

$F^{\prime }(x)=f^{\prime }(x)g(x)+g^{\prime }(x)f(x)$.

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

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

Now we can substitute the values into the equation,

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

Proof

One definition of a derivative is

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

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

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

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

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,

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

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

$F^{\prime }(x)=f^{\prime }(x)g(x)+g^{\prime }(x)f(x)$.