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 \[\begin{aligned} \left.\frac{d}{dx}\sqrt[3]{x}\right|_{x=0} & =\lim_{\Delta x\to0}\frac{\sqrt[3]{0+\Delta x}-\sqrt[3]{0}}{\Delta x}\\ & =\lim_{\Delta x\to0}\frac{\sqrt[3]{\Delta x}}{\Delta x}\\ & =\lim_{\Delta x\to0}\frac{1}{\Delta x^{1-1/3}}\\ & =\lim_{\Delta x\to0}\frac{1}{\sqrt[3]{(\Delta x)^{2}}}\\ & =+\infty.\end{aligned}\]

Figure 3. Tangent to the graph of \(y=\sqrt[3]{x}\) at \((0,0)\) is vertical.

When the increment quotient \({\displaystyle \frac{f(x+\Delta x)-f(x)}{\Delta x}}\) 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 \(y=\begin{cases} x\sin(1/x) & \text{if }x=0\\ 0 & \text{if }x\neq0 \end{cases}\)

 

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