Given $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = |x|^{3/2}$
Then choose the correct option
$1.$ $f$ is differentiable
$2.$ $f$ is differentiable but not continuously differentiable
My attempt: I think option $2$ is correct, i.e., $f$ is differentiable but not continuously differentiable because $f'(x) = 3/2 |x|^{1/2}$ and $f''(x) = 1/2 \frac{1}{\sqrt x}$ not continuous at $x=0$.
Is it true?
The term "continuously differentiable" means that
a. The function is differentiable, and b. The derivative is continuous.
Thus, the second derivative is not relevant here, and the correct option is $1$.
Your assertion $f'(x) = (3/2)|x|^{1/2}$ is certainly wrong. Note: $f$ is decreasing for negative $x$, so $f'(x) < 0$ for negative $x$.
\begin{align} \text{For } x>0, & \qquad \frac d {dx} x^{3/2} = \frac 3 2 x^{1/2} = \frac 3 2 |x|^{1/2}. \\[12pt] \text{For } x<0 \text{ (so that $-x>0$)}, & \qquad \frac d {dx} (-x)^{3/2} = \frac 3 2 (-x)^{1/2} \cdot(-1) \\[12pt] & = - \frac 3 2 |x|^{1/2}. \end{align} So $f(x)$ is a differentiable function of $x$ at points other than $x=0.$
At $x=0$ we have $$ \lim_{\Delta x\,\to\,0} \frac{f(0+\Delta x)-f(0)}{\Delta x} = \lim_{\Delta x\,\to\,0} (\pm |\Delta x|^{3/2}) = 0. $$ Thus $f$ is differentiable at $0.$
The next question is whether $f'$ is continuous. It is at points other than $0.$ At $0$ we see that it approaches $0$ from the left and from the right, and that its value at $0$ is $0.$ Thus it is continuously differentiable everywhere.
