Chapter 3, Problem 2RQ

To determine

Whether the statement, “If f and g are differentiable, then $\frac{d}{dx}\left[f\left(x\right)g\left(x\right)\right]=f^{\prime }\left(x\right)g^{\prime }\left(x\right)$” is true or false.

Answer to Problem 2RQ

The statement is false.

Explanation of Solution

Definition used:

A function $f\left(x\right)$ is differentiable at point x, then $f^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$.

Reason:

Since f is differentiable, then $f^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ (1)

Since g is differentiable, then $g^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{g\left(x+h\right)-g\left(x\right)}{h}$ (2)

Let $F\left(x\right)=f\left(x\right)g\left(x\right)$. Then by definition,

$\begin{array}{c}{F}^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{F\left(x+h\right)-F\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\left[f\left(x+h\right)g\left(x+h\right)\right]-\left[f\left(x\right)g\left(x\right)\right]}{h}\left(QF\left(x\right)=f\left(x\right)g\left(x\right)\right)\\ =\underset{h\to 0}{\lim }\frac{f\left(x+h\right)g\left(x+h\right)-f\left(x+h\right)g\left(x\right)+f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{f\left(x+h\right)\left[g\left(x+h\right)-g\left(x\right)\right]+g\left(x\right)\left[f\left(x+h\right)-f\left(x\right)\right]}{h}\end{array}$

Simplify further,

$\begin{array}{c}{F}^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)\left[g\left(x+h\right)-g\left(x\right)\right]+g\left(x\right)\left[f\left(x+h\right)-f\left(x\right)\right]}{h}\\ =\underset{h\to 0}{\lim }\left[f\left(x+h\right)\right]\underset{h\to 0}{\lim }\frac{\left[g\left(x+h\right)-g\left(x\right)\right]}{h}+\underset{h\to 0}{\lim }\left[g\left(x\right)\right]\underset{h\to 0}{\lim }\frac{\left[f\left(x+h\right)-f\left(x\right)\right]}{h}\\ =f\left(x\right)g^{\prime }\left(x\right)+g\left(x\right)f^{\prime }\left(x\right)\left(\text{by equation (1),(2)}\right)\end{array}$

That is, $\frac{d}{dx}\left[f\left(x\right)g\left(x\right)\right]\ne f^{\prime }\left(x\right)g^{\prime }\left(x\right)$.

Therefore, the given statement is false.