Show that the following functions have no finite

Show that the following functions have no finite derivative at the indicated points: a) $y = \sqrt{x}$ at $x = 0$; b) $y = \sqrt[n]{x-1}$ at $x = 1$ ($n \geq 2$); c) $y = \tan x$ at $x = \dfrac{\pi}{2}+2k\pi$; d) $y = |\sin x|$ at $x = k\pi$, $k = 0,\pm 1,\pm 2,\ldots$.

Step 1: Part (a): $y = \sqrt{x}$ at $x = 0$

The function is defined only for $x \geq 0$, so only a right-hand derivative can be considered. For $\Delta x \gt 0$: \[\frac{\Delta y}{\Delta x} = \frac{\sqrt{\Delta x} - 0}{\Delta x} = \frac{\sqrt{\Delta x}}{\Delta x} = \frac{1}{\sqrt{\Delta x}}.\]

As $\Delta x \to 0^{+}$ this grows without bound: $\lim_{\Delta x \to 0^{+}} \frac{1}{\sqrt{\Delta x}} = +\infty$.

Hence there is no finite derivative at $x = 0$; the graph has a vertical tangent (the $y$-axis) at the origin. $\checkmark$

Step 2: Part (b): $y = \sqrt[n]{x-1}$ at $x = 1$

Here $y(1) = 0$, so for an increment $\Delta x$ the difference quotient is

\[ \frac{\Delta y}{\Delta x} = \frac{\sqrt[n]{(1+\Delta x)-1} - 0}{\Delta x} = \frac{(\Delta x)^{1/n}}{\Delta x} = (\Delta x)^{\frac{1}{n}-1}. \]

From the right ($\Delta x = t \gt 0$), since $\dfrac{n-1}{n} \gt 0$ for $n \geq 2$:

\[ \frac{\Delta y}{\Delta x} = \frac{1}{t^{\,(n-1)/n}} \;\xrightarrow[t\to 0^{+}]{}\; +\infty. \]

From the left the root is defined only for odd $n$. Putting $\Delta x = -t$ with $t \gt 0$ and using $(-t)^{1/n} = -t^{1/n}$ for odd $n$:

\[ \frac{\Delta y}{\Delta x} = \frac{-t^{1/n}}{-t} = t^{\frac{1}{n}-1} = \frac{1}{t^{\,(n-1)/n}} \;\xrightarrow[t\to 0^{+}]{}\; +\infty. \]

So the quotient is unbounded from either side (and only the right side exists for even $n$); no finite derivative exists at $x = 1$. Again the tangent is vertical. $\checkmark$

Step 3: Part (c): $y = \tan x$ at $x_{0} = \dfrac{\pi}{2}+2k\pi$

At such a point $\cos x_{0} = 0$, so $\tan x_{0}$ is undefined. Differentiability at a point presupposes that the function has a value there, so the difference quotient

\[ \frac{\tan(x_{0}+\Delta x)-\tan x_{0}}{\Delta x} \]

cannot even be written down. The behaviour near $x_0$ confirms that no finite value could be assigned: $\lim_{x\to x_{0}^{-}} \tan x = +\infty, \qquad \lim_{x\to x_{0}^{+}} \tan x = -\infty$.

The line $x = x_{0}$ is a vertical asymptote, and no finite derivative exists there. $\checkmark$

Step 4: Part (d): $y = |\sin x|$ at $x_{0} = k\pi$

Here $\sin(k\pi) = 0$, so $f(x_{0}) = 0$. Since $\sin(k\pi+\Delta x) = \pm\sin\Delta x$, taking absolute values gives $\bigl|\sin(k\pi+\Delta x)\bigr| = |\sin\Delta x|$, and therefore

\[ \frac{f(x_{0}+\Delta x)-f(x_{0})}{\Delta x} = \frac{|\sin\Delta x|}{\Delta x}. \]

Using the standard limit $\displaystyle\lim_{u\to 0}\frac{|\sin u|}{|u|} = 1$, the one-sided limits are

\[ f'_{+}(k\pi) = \lim_{\Delta x\to 0^{+}}\frac{\sin\Delta x}{\Delta x} = +1, \qquad f'_{-}(k\pi) = \lim_{\Delta x\to 0^{-}}\frac{-\sin\Delta x}{\Delta x} = -1. \]

Both one-sided derivatives are finite but unequal, so no two-sided derivative exists: the graph has a corner at each zero of the sine, exactly as $|x|$ does at the origin (Problem 20). $\checkmark$

Step 5: Summary of the four mechanisms

\[\text{(a), (b)}\quad \text{the quotient tends to } \infty: \text{ a vertical tangent}\]

\[\text{(c)}\quad \text{the function is undefined at the point: no quotient can be formed}\]

\[\text{(d)}\quad f'_{-} \neq f'_{+}: \text{ a corner}\]