5.1.E: Problems on Derived Functions in One Variable

Exercise $\PageIndex{1}$

Prove Theorems 4 and $5,$ including $\left(i^{*}\right)$ and $\left(i i^{*}\right) .$ Do it for dot products as well.

Exercise $\PageIndex{2}$

Verify Note 2.

Exercise $\PageIndex{3}$

Verify Example (a).

Exercise $\PageIndex{3'}$

Verify Example (b).

Exercise $\PageIndex{4}$

Prove that if $f$ has finite one-sided derivatives at $p,$ it is continuous at $p$.

Exercise $\PageIndex{5}$

Restate and prove Theorems 2 and 3 for one-sided derivatives.

Exercise $\PageIndex{6}$

Prove that if the functions $f_{i} : E^{1} \rightarrow E^{*}(C)$ are differentiable at $p,$ so is their product, and
$$\left(f_{1} f_{2} \cdots f_{m}\right)^{\prime}=\sum_{i=1}^{m}\left(f_{1} f_{2} \cdots f_{i-1} f_{i}^{\prime} f_{i+1} \cdots f_{m}\right) \text { at } p .$$

Exercise $\PageIndex{7}$

A function $f : E^{1} \rightarrow E$ is said to satisfy a Lipschitz condition $(L)$ of order $\alpha(\alpha>0)$ at $p$ iff
$$(\exists \delta>0)\left(\exists K \in E^{1}\right)\left(\forall x \in G_{\neg p}(\delta)\right) \quad|f(x)-f(p)| \leq K|x-p|^{\alpha} .$$
(i) This implies continuity at $p$ but not conversely; take
$$f(x)=\frac{1}{\ln |x|}, \quad f(0)=0, \quad p=0 .$$
$\text { [Hint: For the converse, start with Problem } 14 \text { (iii) of Chapter } 4, §2 .]$
(ii) $L$ of order $\alpha>1$ implies differentiability at $p,$ with $f^{\prime}(p)=0$.
(iii) Differentiability implies $L$ of order $1,$ but not conversely. (Take
$$f(x)=x \sin \frac{1}{x}, f(0)=0, p=0 ;$$
then even one-sided derivatives fail to exist.)

Exercise $\PageIndex{8}$

Let
$$f(x)=\sin x \text { and } g(x)=\cos x .$$
Show that $f$ and $g$ are differentiable on $E^{1},$ with
$$f^{\prime}(p)=\cos p \text { and } g^{\prime}(p)=-\sin p \text { for each } p \in E^{1} .$$
Hence prove for $n=0,1,2, \ldots$ that
$$f^{(n)}(p)=\sin \left(p+\frac{n \pi}{2}\right) \text { and } g^{(n)}(p)=\cos \left(p+\frac{n \pi}{2}\right) .$$
[Hint: Evaluate $\Delta f$ as in Example (d) of Chapter $4, §8 .$ Then use the continuity of $f$ and the formula
$$\lim _{z \rightarrow 0} \frac{\sin z}{z}=\lim _{z \rightarrow 0} \frac{z}{\sin z}=1 .$$
To prove the latter, note that
$$|\sin z| \leq|z| \leq|\tan z| ,$$
whence
$$1 \leq \frac{z}{\sin z} \leq \frac{1}{|\cos z|} \rightarrow 1 ;$$
similarly for $g$.]

Exercise $\PageIndex{9}$

Prove that if $f$ is differentiable at $p$ then
$$\lim _{x \rightarrow p^{+} \atop y \rightarrow p^{-}} \frac{f(x)-f(y)}{x-y} \text { exists, is finite, and equals } f^{\prime}(p) ;$$
i.e., $(\forall \varepsilon>0)(\exists \delta>0)(\forall x \in(p, p+\delta))(\forall y \in(p-\delta, p))$
$$\left|\frac{f(x)-f(y)}{x-y}-f^{\prime}(p)\right|<\varepsilon .$$
Show, by redefining $f$ at $p,$ that even if the limit exists, $f$ may not be $\text { differentiable (note that the above limit does not involve } f(p)) .$
[Hint: If $y<p<x$ then
$$\begin{aligned}\left|\frac{f(x)-f(y)}{x-y}-f^{\prime}(p)\right| & \leq\left|\frac{f(x)-f(p)}{x-y}-\frac{x-p}{x-y} f^{\prime}(p)\right|+\left|\frac{f(p)-f(y)}{x-y}-\frac{p-y}{x-y} f^{\prime}(p)\right| \\ &\left. \leq\left|\frac{f(x)-f(p)}{x-p}-f^{\prime}(p)\right|+\left|\frac{f(p)-f(y)}{p-y}-f^{\prime}(p)\right| \rightarrow 0 .\right] \end{aligned}$$

Exercise $\PageIndex{10}$

Prove that if $f$ is twice differentiable at $p,$ then
$$f^{\prime \prime}(x)=\lim _{h \rightarrow 0} \frac{f(p+h)-2 f(p)+f(p-h)}{h^{2}} \neq \pm \infty .$$
Does the converse hold (cf. Problem 9)?

Exercise $\PageIndex{11}$

In Example $(\mathrm{c}),$ find the three coordinate equations of the tangent line at $p=\frac{1}{2} \pi .$

Exercise $\PageIndex{12}$

Judging from Figure 22 in §2, discuss the existence, finiteness, and sign of the derivatives (or one-sided derivatives) of $f$ at the points $p_{i}$ indicated.

Exercise $\PageIndex{13}$

Let $f : E^{n} \rightarrow E$ be linear, i.e., such that
$$\left(\forall \overline{x}, \overline{y} \in E^{n}\right)\left(\forall a, b \in E^{1}\right) \quad f(a \overline{x}+b \overline{y})=a f(\overline{x})+b f(\overline{y}) .$$
Prove that if $g : E^{1} \rightarrow E^{n}$ is differentiable at $p,$ so is $h=f \circ g$ and $h^{\prime}(p)=f\left(g^{\prime}(p)\right) .$
[Hint: $f$ is continuous since $f(\overline{x})=\sum_{k=1}^{n} x_{k} f\left(\overline{e}_{k}\right) .$ See Problem 5 in Chapter 3, §§4-6.]