Chapter 3, Problem 31RE

To determine

To calculate: The find all the values of x for which the function is differentiable.

Answer to Problem 31RE

The function is differentiable for all values of x except when $x=0$ .

Explanation of Solution

Given Information: The function is $y=\frac{\sin x}{x}$ .

Calculation:

The given function is,

  $y=\frac{\sin x}{x}$

The derivative of the function y with respect to x is calculated as,

  $\begin{array}{c}y=\frac{\sin x}{x}\\ \frac{dy}{dx}=\frac{d}{dx}\left(\frac{\sin x}{x}\right)\\ =\frac{x\frac{d}{dx}\left(\sin x\right)-\sin x\frac{d}{dx}\left(x\right)}{x^2}\left(\text{Apply Quotient rule}\right)\\ =\frac{x\cos x-\sin x\times 1}{x^2}\left(\because \frac{d}{dx}\left(\sin x\right)=\cos x,\frac{d}{dx}\left(x\right)=1\right)\\ =\frac{x\cos x-\sin x}{x^2}\end{array}$

When $x=0,y$ is undefined.

Therefore, the function is not differentiable $x=0$ .

Conclusion:

The function is differentiable for all values of x except when $x=0$.