Question

Prove: If f and g are differentiable and a is in the real numbers, then (fg)'(a) = f(a)g'(a) + f'(a)g(a) and (f/g)'(a) = (g(a)f'(a)-f(a)g'(a))/g(a)^2.

Answer 1

Let's prove product formula using logarithms. Let $f = uv$ and suppose $u$ and $v$ are positive functions of $x$. Then

$$\ln f = \ln (u \cdot v) = \ln u + \ln v.$$

Differentiating both sides:

$$\frac {1}{f} \frac {d f}{d x} = \frac {1}{u} \frac {d u}{d x} + \frac {1}{v} \frac {d v}{d x}$$

and so, multiplying the left side by $f$, and the right side by $uv$,

$$\frac {d f}{d x} = v \frac {d u}{d x} + u \frac {d v}{d x}.$$

2). Suppose $f(x) = g(x) / h(x)$, where $h(x) \neq 0$ and $g$ and $h$ are differentiable.

$$f ^ {\prime} (x) = \lim _ {\Delta x \rightarrow 0} \frac {f (x + \Delta x) - f (x)}{\Delta x} = \lim _ {\Delta x \rightarrow 0} \frac {\frac {g (x + \Delta x)}{h (x + \Delta x)} - \frac {g (x)}{h (x)}}{\Delta x}$$

We pull out the $1 / \Delta x$ and combine the fractions in the numerator:

$$= \lim _ {\Delta x \rightarrow 0} \frac {1}{\Delta x} \left(\frac {g (x + \Delta x) h (x) - g (x) h (x + \Delta x)}{h (x) h (x + \Delta x)}\right)$$

Adding and subtracting $g(x)h(x)$ in the numerator:

$$= \lim _ {\Delta x \rightarrow 0} \frac {1}{\Delta x} \left(\frac {g (x + \Delta x) h (x) - g (x) h (x) - g (x) h (x + \Delta x) + g (x) h (x)}{h (x) h (x + \Delta x)}\right)$$

We factor this and multiply the $1 / \Delta x$ through the numerator:

$$= \lim _ {\Delta x \rightarrow 0} \frac {\frac {g (x + \Delta x) - g (x)}{\Delta x} h (x) - g (x) \frac {h (x + \Delta x) - h (x)}{\Delta x}}{h (x) h (x + \Delta x)}$$

Now we move the limit through:

$$= \frac {\lim _ {\Delta x \rightarrow 0} \left(\frac {g (x + \Delta x) - g (x)}{\Delta x}\right) h (x) - g (x) \lim _ {\Delta x \rightarrow 0} \left(\frac {h (x + \Delta x) - h (x)}{\Delta x}\right)}{h (x) \lim _ {\Delta x \rightarrow 0} h (x + \Delta x)}$$

By the definition of the derivative, the limits of difference quotients in the numerator are derivatives. The limit in the denominator is $h(x)$ because differentiable functions are continuous. Thus we get:

$$= \frac {g ^ {\prime} (x) h (x) - g (x) h ^ {\prime} (x)}{[h (x) ] ^ {2}}.$$