Chapter 1.6, Problem 74E

A proof of the Product Rule appears below. Provide a justification for each step.

a. $\frac{d}{dx}\left[f\left(x\right)\cdot g\left(x\right)\right]=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)g\left(x+h\right)-f\left(x\right)g\left(x\right)}{h}$

Definition of derivative

b. $=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)g\left(x+h\right)-f\left(x+h\right)g\left(x\right)+f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x\right)}{h}$

Adding and subtracting the same quantify is the same as adding 0.

c. $=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)g\left(x+h\right)-f\left(x+h\right)g\left(x\right)}{h}+\underset{h\to 0}{\lim }\frac{f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x\right)}{h}$

The limit of a sum is the sum of the limits.

d. $=\underset{h\to 0}{\lim }\left[f\left(x+h\right)\cdot \frac{g\left(x+h\right)-g\left(x\right)}{h}\right]+\underset{x\to 0}{\lim }\left[g\left(x\right)\cdot \frac{f\left(x+h\right)-f\left(x\right)}{h}\right]$

e. $=f\left(x\right)\cdot \underset{h\to 0}{\lim }\frac{g\left(x+h\right)-g\left(x\right)}{h}+g\left(x\right)\cdot \underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

The limit of a product is the product of the limit and $\underset{h\to 0}{\lim }f\left(x+h\right)=f\left(x\right)$.

f. $f\left(x\right)\cdot g\text{'}\left(x\right)+g\left(x\right)\cdot f\text{'}\left(x\right)$

Definition of derivative

g. $f\left(x\right)\cdot \left[\frac{d}{dx}g\left(x\right)\right]+g\left(x\right)\cdot \left[\frac{d}{dx}f\left(x\right)\right]$

Using Leibniz notation