Answer
$f(x)$ is not differentiable at $P$
Work Step by Step
Consider $f(x)=x$;
The right-hand derivative of $f(x)$ at $P(0,0)$ is given as:
$\lim\limits_{h\to0^+}\dfrac{f(h)-f(0)}{h}=\lim\limits_{h\to 0^+}\dfrac{h-0}{h}=\lim\limits_{h\to0^+}1=1$ ...(1)
Next consider, $f(x)=x^2$
The left-hand derivative of $f(x)$ at $P(0,0)$ is given as:
$\lim\limits_{h\to 0^-}\dfrac{f(h)-f(0)}{h}=\lim\limits_{h\to 0^-}\dfrac{h^2-0}{h}=\lim\limits_{h\to 0^-}h=0$ ...(2)
From the above equations (1) and (2), we conclude that the left-hand derivative is not equal to the right-hand derivative. This means that $f(x)$ is not differentiable at $P$.
