In this section, we study how $$f'(x_{0})$$ might fail to exist. Here are a number of cases in which $$f$$ does not have a derivative at a point.

When $$f$$ is not continuous at $$x=x_{0}$$. For example, if there is a jump in the graph of $$f$$ at $$x=x_{0}$$, or we have $$\lim_{x\to x_{0}}f(x)=+\infty$$ or $$-\infty$$, the function is not differentiable at the point of discontinuity.
For example, consider
$H(x)=\begin{cases} 1 & \text{if }0\leq x\\ 0 & \text{if }x<0 \end{cases}.$ This function, which is called the Heaviside step function, is not continuous at $$x=0$$; therefore $$H'(0)$$ does not exist. Or consider $$F(x)=1/x^{2}.$$ Because $$\lim_{x\to0}F(x)=+\infty$$, $$F$$ is not differentiable at $$x=0$$. Figure 1. The graphs of these two functions show that they are discontinuous at $$x=0$$, hence not differentiable there.

When $$f’_{+}(x_{0})$$ and $$f’_{-}(x_{0})$$ both exist but $$f’_{+}(x_{0})\neq f’_{-}(x_{0})$$. In this case, there is a corner in the graph of $$f$$. For instance, see Figure 2. Figure 2. The function is not differentiable wherever the graph has a corner or cusp.

When the tangent line is vertical. In this case, $\lim_{\Delta x\to0}\frac{f(x_{0}+\Delta x)-f(x_{0})}{\Delta x}=+\infty\ \text{or\ }-\infty.$ For example, consider $$f(x)=x^{1/3}.$$ As you can see in Figure 3, the tangent to its graph at $$(0,0)$$ is vertical. This  unction does not have a derivative at $$x=0$$, for \begin{aligned} \left.\frac{d}{dx}\sqrt{x}\right|_{x=0} & =\lim_{\Delta x\to0}\frac{\sqrt{0+\Delta x}-\sqrt{0}}{\Delta x}\\ & =\lim_{\Delta x\to0}\frac{\sqrt{\Delta x}}{\Delta x}\\ & =\lim_{\Delta x\to0}\frac{1}{\Delta x^{1-1/3}}\\ & =\lim_{\Delta x\to0}\frac{1}{\sqrt{(\Delta x)^{2}}}\\ & =+\infty.\end{aligned} Figure 3. Tangent to the graph of $$y=\sqrt{x}$$ at $$(0,0)$$ is vertical.

When the increment quotient $$\frac{f(x+\Delta x)-f(x)}{\Delta x}$$ does not tend neither to any number nor to $$\pm\infty$$ as $$\Delta x\to0$$.
For example, consider $f(x)=\begin{cases} x\sin\frac{1}{x} & \text{if }x\neq0\\ 0 & \text{if }x=0 \end{cases}.$ This function is not differentiable (although it is continuous) at $$x=0$$, because \begin{aligned} f'(0) & =\lim_{\Delta x\to0}\frac{f(0+\Delta x)-f(0)}{\Delta x}\\ & =\lim_{\Delta x\to0}\frac{\Delta x\sin\frac{1}{\Delta x}-0}{\Delta x}\\ & =\lim_{\Delta x\to0}\sin\frac{1}{\Delta x}\end{aligned} does not exist. The graph of $$f(x)$$ is shown in Figure 4. Figure 4. Graph of $$y=\begin{cases} x\sin(1/x) & \text{if }x=0\\ 0 & \text{if }x\neq0 \end{cases}$$