Proof of Various Derivative Properties — Question 3

Question 3

Assume that $f$ and $g$ are differentiable at $x=a$. Prove the product rule: $$(fg)'(a)=f'(a)g(a)+f(a)g'(a).$$

Question 3 - Solution

By definition of the derivative, $$(fg)'(a)=\lim_{h\to 0}\frac{f(a+h)g(a+h)-f(a)g(a)}{h}.$$

Add and subtract $f(a+h)g(a)$ in the numerator: $$= \lim_{h\to 0} \frac{f(a+h)g(a+h)-f(a+h)g(a)+f(a+h)g(a)-f(a)g(a)}{h}.$$

Group the terms: $$= \lim_{h\to 0} \left[ \frac{f(a+h)\bigl(g(a+h)-g(a)\bigr)}{h} + \frac{g(a)\bigl(f(a+h)-f(a)\bigr)}{h} \right].$$

Separate the limit: $$= \lim_{h\to 0} f(a+h)\, \lim_{h\to 0}\frac{g(a+h)-g(a)}{h} + g(a)\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}.$$

Since $f$ is differentiable at $a$, it is continuous at $a$, so $$\lim_{h\to 0} f(a+h)=f(a).$$

Also, by differentiability, $$\lim_{h\to 0}\frac{g(a+h)-g(a)}{h}=g'(a), \qquad \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=f'(a).$$

Substitute these limits: $$(fg)'(a)=f(a)g'(a)+g(a)f'(a).$$

Thus, $$(fg)'(a)=f'(a)g(a)+f(a)g'(a).$$