Chapter 2.8, Problem 1PT

True or False:

f(x) = tan x is differentiable at $x=\frac{\pi }{2}$.

To determine

Whether the statement “$f\left(x\right)=\tan x$ is differentiable at $x=\frac{\pi }{2}$” is true or false.

Answer to Problem 1PT

The given statement is $\underset{\_}{\text{false}}$.

Explanation of Solution

The derivative of $y=f\left(x\right)$ with respect to x is $f^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$.

The function is differentiable at x means that $f^{\prime }\left(x\right)$ exists.

Compute the value of $f^{\prime }\left(\frac{\pi }{2}\right)$ as follows.

$f^{\prime }\left(\frac{\pi }{2}\right)=\underset{h\to 0}{\lim }\frac{\tan \left(\frac{\pi }{2}+h\right)-\tan \left(\frac{\pi }{2}\right)}{h}$

Clearly, $\underset{h\to 0}{\lim }\frac{\tan \left(\frac{\pi }{2}+h\right)-\tan \left(\frac{\pi }{2}\right)}{h}$ does not exist as $\tan \left(\frac{\pi }{2}\right)$ is not defined.

Hence, $f^{\prime }\left(x\right)$ does not exists at $x=\frac{\pi }{2}$.

That is, $f\left(x\right)$ is not differentiable at $x=\frac{\pi }{2}$.

Therefore, the statement “$f\left(x\right)=\tan x$ is differentiable at $x=\frac{\pi }{2}$” is $\underset{\_}{\text{false}}$.