How to determine whether this function is differentiable at a point?

Solution 1:

The derivative at $0$ is given by the limit

$$\begin{align} f'(0)&=\lim_{h\to 0}\frac{f(h)-f(0)}{h}\\\\ &=\lim_{h\to 0}\frac{f(h)}{h} \end{align}$$

if this limit exists. If $h>0$, then

$$\begin{align} f'(0)&=\lim_{h\to 0^+}\frac{\frac{h}{1+h}}{h}\\\\ &=1 \end{align}$$

If $h<0$, then

$$\begin{align} f'(0)&=\lim_{h\to 0^-}\frac{h^2}{h}\\\\ &=0 \end{align}$$

The right-side and left-side limits are not equal. Therefore, the derivative at $0$ does not exist.

Solution 2:

If $f(x)$ is differentiable then the derivatives from the left and right must be equal at $x=0$. The derivative of $\dfrac{x}{1+x}$ at $x=0$ is $1$. The derivative of $x^2$ is $0$ at $x=0$. Thus $f(x)$ is not differentiable at $x=0$.