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陶哲轩实分析10.3及习题-Analysis I 10.3

图片 2019年5月23日2020年1月7日christangdt留下评论

Exercise 10.3.1. Prove Proposition 10.3.1.
Proposition 10.3.1 Let $X$ be a subset of $\mathbf R$ , let $x_0\in X$ be a limit point of $X$ , and let $f:X\to \mathbf R$ be a function. If $f$ is monotone increasing and $f$ is differentiable at $x_0$ , then $f'(x)\geq 0$ . If $f$ is monotone decreasing and $f$ is differentiable at $x_0$ , then $f'(x)\leq 0$ .

Solution: Since $f$ is differentiable at $x_0$ , we have

$f'(x_0 )=\lim\limits_{x\to x_0;x\in X-{x_0 }} \dfrac{f(x)-f(x_0)}{x-x_0}$

If $f$ is monotone increasing, then we have

$f(x)\leq f(x_0 ),x<x_0,\quad\quad f(x)\geq f(x_0 ),x>x_0$

Thus in both $x<x_0$ and $x>x_0$ we shall have

$\dfrac{f(x)-f(x_0)}{x-x_0}\geq 0 \implies f'(x_0 )=\lim\limits_{x\to x_0;x\in X-{x_0 } }\dfrac{f(x)-f(x_0)}{x-x_0}\geq 0$

The case when $f$ is monotone decreasing can be similarly proved.

$\blacksquare$

Exercise 10.3.2. Give an example of a function $f:(-1,1)\to\mathbf R$ which is continuous and monotone increasing, but which is not differentiable at 0. Explain why this does not contradict Proposition 10.3.1.

Solution: Define

$f(x)=\begin{cases}x,&x\in (-1,0]\\2x,&x\in (0,1) \end{cases}$

This does not contradict Proposition 10.3.1 since Proposition 10.3.1 requires $f$ to be differentiable at the point (0 in this case).

$\blacksquare$

Exercise 10.3.3. Give an example of a function $f:\mathbf R\to\mathbf R$ which is strictly monotone increasing and differentiable, but whose derivative at 0 is zero. Explain why this does not contradict Proposition 10.3.3.

Solution: Define

$f(x)=x^3,\quad x\in (-1,1)$

Then $f$ is differentiable at 0, $f'(0)=0$ , but $f$ is monotone increasing.
This doesn’t contradict Proposition 10.3.1 since $f'>0$ is a necessary but not sufficient condition for $f$ to be strictly monotone increasing.

$\blacksquare$

Exercise 10.3.4. Prove Proposition 10.3.3.
Proposition 10.3.3. Let $a<b$ , and let $f:[a,b]\to\mathbf R$ be a differentiable function. If $f'(x)>0$ for all $x\in [a,b]$ , then $f$ is strictly monotone increasing. If $f'(x)<0$ for all $x\in [a,b]$ , then $f$ is strictly monotone decreasing.If $f'(x)=0$ for all $x\in [a,b]$ , then $f$ is a constant function.

Solution: For any $x\neq y\in [a,b]$ , without loss of generality we can suppose $x<y$ , then $f|_{[x,y]}$ is continuous and differentiable on $[x,y]$ , thus by mean value theorem, we can find a $c\in (x,y)\subset (a,b)$ such that

$f'(c)=\dfrac{f(y)-f(x)}{y-x}=\dfrac{f(x)-f(y)}{x-y}$

If $f'(x)>0,\forall x\in [a,b]$ , then $f'(c)>0$ and $f(x)<f(y)$ , so $f$ is strictly monotone increasing.
If $f'(x)<0,\forall x\in [a,b]$ , then $f'(c)<0$ and $f(x)>f(y)$ , so $f$ is strictly monotone decreasing.
If $f'(x)=0,\forall x\in [a,b]$ , then $f'(c)=0$ and $f(x)=f(y)$ , so $f$ is a constant function.

$\blacksquare$

Exercise 10.3.5. Give an example of a subset $X\subset \mathbf R$ and a function $f:X\to\mathbf R$ which is differentiable on $X$ , is such that $f'(x)>0$ for all $x\in X$ , but $f$ is not strictly monotone increasing.

Solution: Define

$f(x)=\begin{cases}x+1,&x\in (-1,0)\\x-1,&x\in (0,1)\end{cases}$

Then if $X=(-1,0)\cup (0,1)$ , then $f$ is differentiable on $X$ and $f'(x)=1>0,\forall x\in X$ , but we have $f(1/2)=-1/2<f(-1/2)=1/2$ , thus $f$ in not strictly monotone increasing.
The key condition which is different from Proposition 10.3.3 is that $X$ is allowed to be a disconnected set.

$\blacksquare$

陶哲轩实分析10.2及习题-Analysis I 10.2

2019年5月22日2020年1月7日christangdt留下评论

局部最值，Rolle定理和中值定理（Lagrange）。

Exercise 10.2.1. Prove Proposition 10.2.6.
Proposition 10.2.6 (Local extrema are stationary). Let $a<b$ be real numbers,and let $f:(a,b)\to\mathbf R$ be a function. If $x_0\in (a,b)$ , $f$ is differentiable at $x_0$ ,and $f$ attains either a local maximum or local minimum at $x_0$ , then $f'(x_0)=0$ .

Solution: First suppose $f$ attains a local maximum at $x_0$ , then $\exists \delta >0$ , s.t.

$f(x)\leq f(x_0 ),\quad \forall x\in (a,b)\cap (x_0-\delta ,x_0+\delta )$

Since $f$ is differentiable at $x_0$ , the limit

$\lim\limits_{x\to x_0;x\in (a,b))}\dfrac{f(x)-f(x_0)}{x-x_0}$

exists, thus both left and right limit exists and is equal to $f'(x_0 )$ .
Notice that

$\dfrac{f(x)-f(x_0)}{x-x_0 }\geq 0,\quad x\in (a,b)\cap (x_0-\delta ,x_0 )$

and

$\dfrac{f(x)-f(x_0)}{x-x_0 }\leq 0,\quad x\in (a,b)\cap (x_0,x_0+\delta )$

We shall have

$f'(x_0 )=\lim\limits_{x\to x_0;x\in (a,b)\cap (x_0-\delta ,x_0 ) }\dfrac{f(x)-f(x_0)}{x-x_0 }\geq 0$

and

$f'(x_0 )=\lim\limits_{x\to x_0;x\in (a,b)\cap (x_0,x_0+\delta ) }\dfrac{f(x)-f(x_0)}{x-x_0 }\leq 0$

Thus $f'(x_0 )=0$ .
The case when $f$ attains a local minimum at $x_0$ can be similarly proved.

$\blacksquare$

Exercise 10.2.2. Give an example of a function $f:(-1,1)\to\mathbf R$ which is continuous and attains a golbal maximum at 0, but which is not differentiable at 0. Explain why this does not contradict Proposition 10.2.6.

Solution: Define

$f(x)=-|x|,\quad x\in (-1,1)$

Then $f$ is continuous and attains a global maximum at $0$ , but is not differentiable at $0$ .
This doesn’t contradict Proposition 10.2.6, since Proposition 10.2.6 requires $f$ to be differentiable at local maximum or minimum.

$\blacksquare$

Exercise 10.2.3. Give an example of a function $f:(-1,1)\to\mathbf R$ which is differentiable, and whose derivative equals 0 at 0, but such that 0 is neither a local minimum nor a local maximum. Explain why this does not contradict Proposition 10.2.6.

Solution: Define

$f(x)=x^3,\quad x\in (-1,1)$

Then $f$ is differentiable at 0, $f'(0)=0$ , but $0$ is neither a local minimum nor local maximum.
This doesn’t contradict Proposition 10.2.6, since derivative equals zero is a necessary but not sufficient condition for a a local minimum nor local maximum.

$\blacksquare$

Exercise 10.2.4. Prove Theorem 10.2.7.
Theorem 10.2.7 (Rolle’s theorem). Let $a<b$ be real numbers, and let $g:[a,b]\to\mathbf R$ be a continuous function which is differentiable on $(a,b)$ . Shppose also that $g(a)=g(b)$ . Then there exists an $x\in (a,b)$ such that $g'(x)=0$ .

Solution: If $\forall x\in (a,b),g(x)=g(a)=g(b)$ , then $g$ is constant on $[a,b]$ and $g'(x)=0,x\in (a,b)$ by Theorem 10.1.13.
Now suppose if $\exists x\in (a,b),g(x)\neq g(a)$ , then $g(x)\neq g(b)$ . Since $g$ is continuous on $[a,b]$ we can use the maximum principle, say $f$ attains its maximum at $x_{max}\in [a,b]$ , attains it’s minimum at $x_{min}\in [a,b]$ . We either have $g(x)>g(a)$ or $g(x)<g(a)$ by the trichotomy of real numbers.
If $g(x)>g(a)$ , then $f$ cannot attain its maximum at a or b, so $x_{max}\in (a,b)$ , since $x_{max}$ is also a local maximum, we have $f'(x_{max})=0$ by Proposition 10.2.6.
If $g(x)<g(a)$ , then $f$ can’t attain its minimum at $a$ or $b$ , so $x_{min}\in (a,b)$ , since $x_{min}$ is also a local minimum, we have $f'(x_{min} )=0$ by Proposition 10.2.6.
In conclusion the statement is true in all cases.

$\blacksquare$

Exercise 10.2.5. Use Theorem 10.2.7 to prove Corollary 10.2.9.
Corollary 10.2.9 (Mean value theorem). Let $a<b$ be real numbers,and let $f:(a,b)\to\mathbf R$ be a function which is continuous on $[a,b]$ and differentiable on $(a,b)$ . Then there exists an $x\in (a,b)$ such that $f'(x)=\frac{f(b)-f(a)}{b-a}$ .

Solution: We let

$F(y)=f(y)-\dfrac{f(b)-f(a)}{b-a} (y-a)$

Then by Proposition 9.4.9 and Theorem 10.1.13, $F(y)$ is continuous on $[a,b]$ and differentiable on $(a,b)$ . Since $F(a)=f(a),F(b)=f(a)$ , we can use Rolle’s theorem to say $\exists x\in (a,b),F'(x)=0$ , or

$F'(x)=f'(x)-\dfrac{f(b)-f(a)}{b-a} =0 \implies f'(x)=\dfrac{f(b)-f(a)}{b-a}$

$\blacksquare$

Exercise 10.2.6. Let $M>0$ , and let $f:[a,b]:\to\mathbf R$ be a function which is continuous on $[a,b]$ and differentiable on $(a,b)$ , and such that $|f'(x)|\leq M$ for all $x\in (a,b)$ (i.e.,the derivative of $f$ is bounded). Show that for any $x,y\in [a,b]$ we have the inequality $|f(X)-f(y)|\leq M|x-y|$ . Functions which obey the bound $|f(X)-f(y)|\leq M|x-y|$ are known as Lipschitz continuous functions with Lipschitz constant M; thus this exercise shows that functions with bounded derivative are Lipschitz continuous.

Solution: For any $x,y\in [a,b]$ , if $x=y$ the statement is obviously true. Without loss of generality we can suppose $x<y$ , then $f|_{[x,y]}$ is continuous on $[x,y]$ and differentiable on $(x,y)$ , thus by mean value theorem, we can find a $c\in (x,y)\subset (a,b)$ such that

$f'(c)=\dfrac{f(y)-f(x)}{y-x}=\dfrac{f(x)-f(y)}{x-y}$

Since $|f'(x)|\leq M,\forall x\in (a,b)$ , we have

$|f'(c)|=\left|\dfrac{f(x)-f(y)}{x-y}\right|\leq M\implies |f(x)-f(y)|\leq M|x-y|$

$\blacksquare$

Exercise 10.2.7. Let $f:\mathbf R\to\mathbf R$ be a differentiable function such that $f'$ is bounded. Show that $f$ is uniformly continuous.

Solution: Since $f'$ is bounded, $\exists M>0$ , s.t. $|f'(x)|\leq M,\forall x\in \mathbf R$ . Then given $\forall \varepsilon >0$ , let $\delta =\varepsilon /M$ , then as long as $|x-y|<\delta$ , we can get from Exercise 10.2.6 that

$|f(x)-f(y)|\leq M|x-y|<M\delta =\varepsilon$

which means $f$ is uniformly continuous.

$\blacksquare$

陶哲轩实分析10.1及习题-Analysis I 10.1

2019年5月22日2021年8月14日christangdt12条评论

介绍了微分的许多基本概念。

Exercise 10.1.1. Suppose that $X$ is a subset of $\mathbf R$ , $x_0$ is a limit point of $X$ , and $f:X\to\mathbf R$ is a function which is differentiable at $x_0$ . Let $Y\subset X$ be such that $x_0 \in Y$ , and $x_0$ is also a limit point of $Y$ . Prove that the restricted function $f|_Y:Y\to\mathbf R$ is also differentiable at $x_0$ , and has the same derivative as $f$ at $x_0$ . Explain why this does not contradict the discussion in Remark 10.1.2.

Solution: $f$ is differentiable at $x_0$ means the limit

$\lim\limits_{x\to x_0;x\in X-\{x_0 \}}\dfrac{f(x)-f(x_0)}{x-x_0}$

exists, suppose the limit is $L$ , then by definition this means $\forall \varepsilon >0,\exists \delta >0,$ s.t.

$\left|\dfrac{f(x)-f(x_0 )}{x-x_0 }-L\right|<\varepsilon ,\quad\forall x\in \Big((x_0-\delta ,x_0+\delta )\cap X\Big)-\{x_0 \}$

Consider $f|_Y$ , if $\forall y\in (x_0-\delta ,x_0+\delta )\cap Y$ and $y\neq x_0$ , then since $Y\subset X$ , we have $y\in \Big((x_0-\delta ,x_0+\delta )\cap X\Big)-\{x_0 \}$ , so

$\left|\dfrac{f|_Y (y)-f|_Y (x_0 )}{y-x_0 }-L\right|=\left|\dfrac{f(y)-f(x_0 )}{y-x_0 }-L\right|<\varepsilon$

which means the limit $\lim_{y\to x_0;y\in Y-\{x_0 \} }\frac{f|_Y (y)-f|_Y (x_0 )}{y-x_0}$ exists, i.e., $f|_Y$ is differentiable at $x_0$ .

$\blacksquare$

Exercise 10.1.2. Prove Proposition 10.1.7.
Proposition 10.1.7 (Newton’s approximation). Let $X$ be a subset of $\mathbf R$ , let $x_0\in X$ be a limit point of $X$ , let $f:X\to\mathbf R$ be a function, and let $L$ be a real number. Then the following statements are logically equivalent:
( a ) $f$ is differentiable at $x_0$ on $X$ with derivative $L$ .
( b ) For every $\varepsilon >0$ , there exists a $\delta >0$ such that $f(x)$ is $\varepsilon |x-x_0|$ -close to $f(x)+L(x-x_0)$ whenever $x\in X$ is $\delta$ -close to $x_0$ , i.e.,we have

$|f(x)-(f(x_0)+L(x-x_0))|\leq \varepsilon |x-x_0|$

whenever $x\in X$ and $|x-x_0|\leq \delta$ .

Solution: (a) implies (b):
If (a) is valid, then

$\lim\limits_{x\to x_0;x\in X-\{x_0 \}}\dfrac{f(x)-f(x_0)}{x-x_0 }=L$

Thus $\forall \varepsilon >0,\exists \delta >0$ , s.t. whenever $x\in ([x_0-\delta ,x_0+\delta ]\cap X)-\{x_0 \}$

$\left|\dfrac{f(x)-f(x_0)}{x-x_0 }-L\right|\leq \varepsilon \implies |f(x)-f(x_0 )-L(x-x_0 )|\leq \varepsilon |x-x_0 |$

Note that if $x=x_0$ , then $|f(x)-f(x_0 )-L(x-x_0 )|=0\leq 0=\varepsilon |x-x_0 |$ . So we can conclude (b) is true.
(b) implies (a):
If (b) is true, then for every $\varepsilon >0$ , we let $x\in ([x_0-\delta ,x_0+\delta ]\cap X)-\{x_0 \}$ , obviously $x$ satisfies the condition of (b) and in addition we have $|x-x_0 |\neq 0$ , thus we have

$|f(x)-(f(x_0 )+L(x-x_0 ))|\leq \varepsilon |x-x_0 |$

And from this we can further have

$\left|\dfrac{f(x)-f(x_0)}{x-x_0 }-L\right|\leq \varepsilon$

which means (a) is true since

$\lim\limits_{x\to x_0;x\in X-\{x_0 \}} \dfrac{f(x)-f(x_0)}{x-x_0 }=L$

$\blacksquare$

Exercise 10.1.3. Prove Proposition 10.1.10.
Proposition 10.1.10 (Differentiability implies continuity). Let $X$ be a subset of $\mathbf R$ , let $x_0\in X$ be a limit point of $X$ , and let $f:X\to\mathbf R$ be a function. If $f$ is differentiable at $x_0$ , then $f$ is also continuous at $x_0$ .

Solution: If $f$ is differentiable at $x_0$ and we let the derivative be $L$ , obviously $|L|<+{\infty}$ , by Proposition 10.1.7 we can have $\forall \varepsilon >0,\exists \delta >0$ , s.t.

$|f(x)-(f(x_0 )+L(x-x_0 ))|\leq \varepsilon |x-x_0 |,\quad \forall x\in X,|x-x_0 |\leq \delta$

This means

$|f(x)-f(x_0 )|\leq (L+\varepsilon )|x-x_0 |,\quad \forall x\in X,|x-x_0 |\leq \delta$

Now let $\delta '=\min (\delta ,\varepsilon /(L+\varepsilon ))$ , then if $x\in X,|x-x_0 |\leq \delta '$ , we have $|f(x)-f(x_0 )|\leq \varepsilon$ , so $f$ is continuous at $x_0$ .
If instead we want to use the limit laws, we can see that

$\begin{aligned} \lim\limits_{x\to x_0;x\in X-\{x_0 \}} \dfrac{f(x)-f(x_0)}{x-x_0}=L &\implies \lim\limits_{x\to x_0;x\in X-\{x_0 \}}(f(x)-f(x_0 ))=\lim\limits_{x\to x_0;x\in X-\{x_0 \} } L(x-x_0 ) \\&\implies \lim\limits_{x\to x_0} (f(x)-f(x_0))=\lim\limits_{x\to x_0}L(x-x_0 )=L \lim\limits_{x\to x_0} (x-x_0 )=0 \\&\implies \lim\limits_{x\to x_0}f(x)=\lim\limits_{x\to x_0 } f(x_0)=f(x_0 )\end{aligned}$

$\blacksquare$

Exercise 10.1.4. Prove Theorem 10.1.13.
Theorem 10.1.13(Differential calculus). Let $X$ be a subset of $\mathbf R$ , let $x_0\in X$ be a limit point of $X$ , and let $f:X\to\mathbf R$ and $g:X\to\mathbf R$ be functions.

( a ) If $f$ is a constant function, i.e., there exists a real number $c$ such that $f(x)=c$ for all $x\in X$ , then $f$ is differentiable at $x_0$ and $f'(x_0)=0$ .
Solution: We have

$\lim\limits_{x\to x_0;x\in X-\{x_0 \}} \dfrac{f(x)-f(x_0)}{x-x_0}=\lim\limits_{x\to x_0;x\in X-\{x_0 \}} \dfrac{c-c_0}{x-x_0}=\lim\limits_{x\to x_0;x\in X-\{x_0 \}} 0=0$

( b ) If $f$ is the identity function, i.e., $f(x)=x$ for all $x\in X$ , then $f$ is differentiable at $x_0$ and $f'(x_0)=1$ .
Solution: We have

$\lim\limits_{x\to x_0;x\in X-\{x_0 \}} \dfrac{f(x)-f(x_0)}{x-x_0}=\lim\limits_{x\to x_0;x\in X-\{x_0 \}} \dfrac{x-x_0}{x-x_0}=\lim\limits_{x\to x_0;x\in X-\{x_0 \}} 1=1$

( c ) (Sum rule) If $f$ and $g$ are differentiable at $x_0$ , then $f+g$ is also differentiable at $x_0$ , and $(f+g)'(x_0)=f'(x_0)+g'(x_0)$ .
Solution: We have

$\lim\limits_{x\to x_0;x\in X-\{x_0 \}} \dfrac{(f+g)(x)-(f+g)(x_0)}{x-x_0}=\lim\limits_{x\to x_0;x\in X-\{x_0 \}} \dfrac{f(x)+g(x)-f(x_0 )-g(x_0)}{x-x_0 }\\=\lim\limits_{x\to x_0;x\in X-\{x_0 \}}\dfrac{f(x)-f(x_0)}{x-x_0}+\lim\limits_{x\to x_0;x\in X-\{x_0 \}} \dfrac{g(x)-g(x_0)}{x-x_0 }=f'(x_0 )+g'(x_0 )$

Thus $f+g$ is differentiable at $x_0$ and $(f+g)'(x_0 )=f'(x_0 )+g'(x_0 )$

( d ) (Product rule) If $f$ and $g$ are differentiable at $x_0$ , then $fg$ is also differentiable at $x_0$ , and $(fg)'(x_0)=f'(x_0)g(x_0)+f(x_0)g'(x_0)$ .
Solution: We have, use the “middle-man trick”, that

$\lim\limits_{x\to x_0;x\in X-\{x_0 \}}\dfrac{(fg)(x)-(fg)(x_0)}{x-x_0}=\lim\limits_{x\to x_0;x\in X-\{x_0 \}}\dfrac{f(x)(g(x)-g(x_0 ))+(f(x)-f(x_0 ))g(x_0 )}{x-x_0}\\=\lim\limits_{x\to x_0;x\in X-\{x_0 \}}\dfrac{f(x)(g(x)-g(x_0)}{x-x_0}+\lim\limits_{x\to x_0;x\in X-\{x_0 \}}\dfrac{f(x)-f(x_0 )}{x-x_0} g(x_0 )$

Use Proposition 10.1.10 and limit laws, we can further calculate:

$\begin{aligned}&{\ }{\ }{\ }{\ }{\ }\lim\limits_{x\to x_0;x\in X-\{x_0 \}}\dfrac{f(x)(g(x)-g(x_0)}{x-x_0}+\lim\limits_{x\to x_0;x\in X-\{x_0 \}}\dfrac{f(x)-f(x_0 )}{x-x_0} g(x_0 )\\&=\left(\lim\limits_{x\to x_0;x\in X-\{x_0 \}} f(x) \right)\left(\lim\limits_{x\to x_0;x\in X-\{x_0 \}}\dfrac{g(x)-g(x_0 )}{x-x_0 }\right)+g(x_0 ) \lim\limits_{x\to x_0;x\in X-\{x_0 \}}\dfrac{f(x)-f(x_0 )}{x-x_0 }\\&=f(x_0 ) g'(x_0 )+f'(x_0 )g(x_0 )\end{aligned}$

( e ) If $f$ is differentiable at $x_0$ and $c$ is a real number, then $cf$ is also differentiable at $x_0$ , and $(cf)'(x_0)=cf'(x_0)$ .
Solution: Combining (a) and (d), let $g(x)=c$ , then $g'(x)=0$ , and $cf$ differentiable at $x_0$ , and

$(cf)'(x_0 )=(gf)'(x_0 )=f(x_0 ) g'(x_0 )+f'(x_0 )g(x_0 )=cf'(x_0 )$

( f ) (Difference rule) If $f$ and $g$ are differentiable at $x_0$ , then $f-g$ is also differentiable at $x_0$ , and $(f-g)'(x_0)=f'(x_0)-g'(x_0)$ .
Solution: from (e) we know $-g$ is differentiable at $x_0$ , $(-g)'(x_0 )=-g'(x_0 )$ , thus by (a) we have

$(f-g)'(x_0 )=\left(f+(-g))'(x_0 \right)=f'(x_0 )+(-g)'(x_0 )=f'(x_0 )-g'(x_0 )$

( g ) If $g$ is differentiable at $x_0$ , and $g$ is non-zero on $X$ (i.e., $g(x)\neq 0$ for all $x\in X$ ), then $1/g$ is also differentiable at $x_0$ , and $\left(\frac{1}{g}\right)'(x_0)=-\frac{g'(x_0)}{g(x_0)^2}$
Solution: We have, use Proposition 10.1.10 and limit laws,

$\lim\limits_{x\to x_0;x\in X-\{x_0 \}}\dfrac{(1/g)(x)-(1/g)(x_0)}{x-x_0 }=\lim\limits_{x\to x_0;x\in X-\{x_0 \}}\dfrac{1/g(x)-1/g(x_0)}{x-x_0}\\=\lim_{x\to x_0;x\in X-\{x_0 \}}\dfrac{-1}{(g(x)g(x_0)} \dfrac {g(x)-g(x_0)}{x-x_0}=-\dfrac{g'(x_0 )}{g(x_0 )^2 }$

( h ) (Quotient rule) If $f$ and $g$ are differentiable at $x_0$ , and $g$ is non-zero on $X$ , then $f/g$ is also differentiable at $x_0$ , and

$(\dfrac{f}{g})'(x_0)=\dfrac{f'(x_0)g(x_0)-f(x_0)g'(x_0)}{g(x_0)^2}$

Solution: by (d) and (g), $f/g$ is differentiable at $x_0$ , and

$\begin{aligned}\left(\dfrac{f}{g}\right)'(x_0 )&=\left(f\left(\dfrac{1}{g}\right)\right)'(x_0 )=f'(x_0 )\left(\dfrac{1}{g}\right)(x_0 )+f(x_0 ) \left(\dfrac{1}{g}\right)'(x_0 )\\&=\dfrac{f'(x_0 )}{g(x_0 )} -f(x_0 ) \dfrac{g'(x_0 )}{g(x_0 )^2 }=\dfrac{f'(x_0 )g(x_0 )-f(x_0 ) g'(x_0 )}{(g(x_0 )^2 }\end{aligned}$

$\blacksquare$

Exercise 10.1.5. Let $n$ be a natural number, and let $f:\mathbf R\to\mathbf R$ be the function $f(x):=x^n$ . Show that $f$ is differentiable on $\mathbf R$ and $f'(x)=nx^{n-1}$ for all $x\in \mathbf R$ .

Solution: If $n=0$ , then $f(x)=x^0=1$ , and $f'(x)=0=0x^(-1)=0,x\in \mathbf R-\{0\}$ , when $x=0$ , use definition we can show $f'(0)=0$ . Thus $f$ is differentiable on $\mathbf R$ .
Now suppose the case is true for $n$ , then in the case of $n+1$ , we use Theorem 10.1.13 and have $f(x)=x^{n+1}=x^n\cdot x$ is differentiable on $\mathbf R$ , and

$f'(x)=(x^n )'x+(x^n ) (x)'=nx^n+x^n=(n+1) x^n,\quad \forall x\in \mathbf R$

Thus the induction hypothesis holds.

$\blacksquare$

Exercise 10.1.6. Let $n$ be a negative integer, and let $f:\mathbf R-\{0\}\to\mathbf R$ be the function $f(x):=x^n$ . Show that $f$ is differentiable on $\mathbf R$ and $f'(x)=nx^{n-1}$ for all $x\in \mathbf R-\{0\}$ .

Solution: Use negative induction, first let $n=-1$ , so $f(x)=x^{-1}$ , then $f'(x)=-x^{-2}$ , the induction hypothesis is satisfied.
Now suppose for negative integer $n$ the induction holds, let $f(x)=x^{n-1}=x^n\cdot x^{-1}$ , by Theorem 10.1.13 $f(x)$ is differentiable on $\mathbf R-\{0\}$ , and

$f'(x)=(x^n )'(x^{-1} )+x^n (x^{-1} )'=nx^{n-2}-x^{n-2}=(n-1) x^{n-2}$

Thus the induction hypothesis holds.

$\blacksquare$

Exercise 10.1.7. Prove Theorem 10.1.15.
Theorem 10.1.15(Chain rule). Let $X,Y$ be subsets of $\mathbf R$ , let $x_0\in X$ be a limit point of $X$ , and let $y_0\in Y$ be a limit point of $Y$ . Let $f:X\to Y$ be a function such that $f(x_0)=y_0$ , and such that $f$ is differentiable at $x_0$ . Suppose that $g:Y\to\mathbf R$ is a function which is differentiable at $y_0$ . Then the function $g\circ f:X\to\mathbf R$ is differentiable at $x_0$ , and

$(g\circ f)'(x_0)=g'(y_0)f'(x_0)$

Solution: We have $f'(x_0 )=L_1$ and $g'(y_0 )=L_2$ exists and is finite. Then for $\forall \varepsilon >0, \exists \varepsilon '>0$ s.t.

$\varepsilon '^2+|L_1 | \varepsilon '+|L_2 | \varepsilon '\leq \varepsilon$

For this $\varepsilon '$ , use Proposition 10.1.7, $\exists \delta _1>0$ , s.t.

$|g(y)-g(y_0 )-L_2 (y-y_0)|\leq \varepsilon '|y-y_0 |,\quad \forall y\in Y,|y-y_0 |\leq \delta _1$

For this $\delta _1,\exists \delta _2>0$ , s.t.

$|f(x)-f(x_0)|<\delta _1,\quad \forall x\in X,|x-x_0 |<\delta _2$

We can also have a $\delta _3>0$ , s.t.

$|f(x)-f(x_0 )-L_1 (x-x_0)|\leq \varepsilon '|x-x_0 |,\quad \forall x\in X,|x-x_0 |\leq \delta_3$

Let $\delta =\min (\delta _2,\delta _3)$ , then when $x\in X,|x-x_0 |\leq \delta$ , we shall have $|f(x)-f(x_0)|<\delta _1$ , and

$|g(f(x))-g(f(x_0 ))-L_2 (f(x)-f(x_0 ))|\leq \varepsilon '|f(x)-f(x_0)|$

Use the triangle inequality we can further deduce

$\begin{aligned}&{\ }{\ }{\ }{\ }|g(f(x))-g(f(x_0 ))-L_2 (f(x)-f(x_0 ))|\\&=|g(f(x))-g(f(x_0 ))-L_2 (f(x)-f(x_0 )-L_1 (x-x_0 ))-L_2 L_1 (x-x_0)|\\&\geq |g(f(x))-g(f(x_0 ))-L_2 L_1 (x-x_0 )|-|L_2 (f(x)-f(x_0 )-L_1 (x-x_0 ))|\end{aligned}$

Thus

$\begin{aligned}&{\ }{\ }{\ }{\ }|g(f(x))-g(f(x_0 ))-L_2 L_1 (x-x_0 )|\\&\leq |g(f(x))-g(f(x_0 ))-L_2 (f(x)-f(x_0 ))|+|L_2 (f(x)-f(x_0 )-L_1 (x-x_0 ))|\\&\leq \varepsilon '|f(x)-f(x_0 )|+|L_2 | \varepsilon '|x-x_0 |\\&\leq \varepsilon '^2 |x-x_0 |+|L_1 | \varepsilon '|x-x_0 |+|L_2 | \varepsilon '|x-x_0 |\\&=(\varepsilon '^2+|L_1 | \varepsilon '+|L_2 | \varepsilon ')|x-x_0 |\leq \varepsilon |x-x_0 |\end{aligned}$

Use proposition 10.1.7, we can say the function $g\circ f$ is differentiable at $x_0$ , and

$(g\circ f)'(x_0 )=L_2 L_1=g'(y_0 ) f'(x_0 )$

$\blacksquare$

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