# 10.5 The product rule

Proving this is a little tricky, and needs some ingenuity. The product rule gives us a way to find the derivative of a function which is the product of two functions $f(x)=a(x)\cdot b(x)$.

The trick here is to "add zero"

$\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ | $\displaystyle=\frac{a(x+h)b(x+h)-a(x)b(x)}{h}$ | (10.27) | ||

$\displaystyle=\lim_{h\to 0}\frac{a(x+h)b(x+h)-a(x+h)b(x)+a(x+h)b(x)-a(x)b(x)}{h}$ | (10.28) | |||

$\displaystyle=\lim_{h\to 0}\frac{a(x+h)(b(x+h)-b(x))+b(x)(a(x+h)-a(x)}{h}$ | (10.29) | |||

$\displaystyle=\lim_{h\to 0}\frac{a(x+h)(b(x+h)-b(x))}{h}+\lim_{h\to 0}\frac{b(% x)(a(x+h)-a(x)}{h}$ | (10.30) | |||

$\displaystyle=\lim_{h\to 0}a(x+h)\frac{(b(x+h)-b(x))}{h}+\lim_{h\to 0}b(x)% \frac{(a(x+h)-a(x)}{h}$ | (10.31) |

If (as it does) $h\to 0$ then $a(x+h)\to a(x)$, we can rewrite the limit as

$\displaystyle a(x)\lim_{h\to 0}\frac{b(x+h)-b(x)}{h}+b(x)\lim_{h\to 0}\frac{a(% x+h)-a(x)}{h}$ | $\displaystyle=a(x)\frac{d}{dx}[b(x)]+b(x)\frac{d}{dx}[a(x)]+b(x)$ | (10.32) |

Overall, we therefore can say that the derivative of a function $f(x)=a(x)b(x)$ is

$\frac{df}{dx}=\frac{d}{dx}[a(x)]b(x)+a(x)\frac{d}{dx}[b(x]$
(10.33)