Proof of Various Derivative Properties — Question 4

Question 4

Assume that $f$ and $g$ are differentiable at $x=a$ and that $g(a)\neq 0$. Prove the quotient rule: $$\left(\frac{f}{g}\right)'(a) = \frac{f'(a)g(a)-f(a)g'(a)}{[g(a)]^2}.$$

Question 4 - Solution

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

Combine the fractions in the numerator: $$= \lim_{h\to 0} \frac{f(a+h)g(a)-f(a)g(a+h)}{h\,g(a+h)g(a)}.$$

Rewrite the numerator by adding and subtracting $f(a)g(a)$: $$f(a+h)g(a)-f(a)g(a+h) = g(a)\bigl(f(a+h)-f(a)\bigr) - f(a)\bigl(g(a+h)-g(a)\bigr).$$

Substitute this expression: $$= \lim_{h\to 0} \frac{ g(a)\bigl(f(a+h)-f(a)\bigr) - f(a)\bigl(g(a+h)-g(a)\bigr) }{ h\,g(a+h)g(a) }.$$

Separate the fraction: $$= \lim_{h\to 0} \left[ \frac{g(a)}{g(a+h)g(a)} \cdot \frac{f(a+h)-f(a)}{h} - \frac{f(a)}{g(a+h)g(a)} \cdot \frac{g(a+h)-g(a)}{h} \right].$$

Simplify coefficients: $$= \lim_{h\to 0} \left[ \frac{1}{g(a+h)} \cdot \frac{f(a+h)-f(a)}{h} - \frac{f(a)}{g(a)g(a+h)} \cdot \frac{g(a+h)-g(a)}{h} \right].$$

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

Taking limits term by term, $$\left(\frac{f}{g}\right)'(a) = \frac{1}{g(a)}f'(a) - \frac{f(a)}{[g(a)]^2}g'(a).$$

Combine the terms: $$\left(\frac{f}{g}\right)'(a) = \frac{f'(a)g(a)-f(a)g'(a)}{[g(a)]^2}.$$