MATH 221

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 4 August 2014

Lecture 9

Leftovers: ${\text{log}}_{a}x$ is the inverse function to $a^x\text{.}$ So $${\text{log}}_a{a}^{\sqrt{7}\pi i\text{sin}\hspace{0.17em}32}=\sqrt{7}\pi i\text{sin}\hspace{0.17em}32$$ and $$a^{{\text{log}}_a\left(\sqrt{7}\pi i\text{sin}\hspace{0.17em}32\right)}=\sqrt{7}\pi i\text{sin}\hspace{0.17em}32\text{.}$$

Find $\frac{dy}{dx}$ when $y={\text{log}}_{x}10\text{.}$ So $$x^y=x^{{\text{log}}_{x}10}=10\text{.}$$ Take the derivative: $$\frac{d{x}^y}{dx}=\frac{d{\left(e^{\text{ln}\hspace{0.17em}x}\right)}^y}{dx}=\frac{d{e}^{y\text{ln}\hspace{0.17em}x}}{dx}=e^{y\text{ln}\hspace{0.17em}x}(y\frac{1}{x}+\frac{dy}{dx}\text{ln}\hspace{0.17em}x)\text{.}$$ So $$e^{y\text{ln}\hspace{0.17em}x}(\frac{y}{x}+\frac{dy}{dx}\text{ln}\hspace{0.17em}x)=0\text{.}$$ (Since $x^y=10$ and $\frac{d10}{dx}=0\text{.)}$ Solve for $\frac{dy}{dx}\text{.}$ $$\begin{array}{ccc}{\displaystyle e^{y\text{ln}\hspace{0.17em}x}\frac{dy}{dx}\text{ln}\hspace{0.17em}x}& =& {\displaystyle -\frac{e^{y\text{ln}\hspace{0.17em}x}y}{x}}\\ {\displaystyle \frac{dy}{dx}}& =& {\displaystyle -\frac{e^{y\text{ln}\hspace{0.17em}x}y}{x{e}^{y\text{ln}\hspace{0.17em}x}\text{ln}\hspace{0.17em}x}=-\frac{y}{x\text{ln}\hspace{0.17em}x}=-\frac{{\text{log}}_{x}10}{x\text{ln}\hspace{0.17em}x}\text{.}}\end{array}$$

Find the third derivative of $2^x$ with respect to $x\text{.}$ $$\begin{array}{ccc}{\displaystyle y}& =& {\displaystyle 2^x}\\ {\displaystyle \frac{dy}{dx}}& =& {\displaystyle \frac{d{2}^x}{dx}=\frac{d{\left(e^{\text{ln}\hspace{0.17em}2}\right)}^x}{dx}=\frac{d{e}^{x\text{ln}\hspace{0.17em}2}}{dx}}\\ & =& {\displaystyle e^{x\text{ln}\hspace{0.17em}2}\left(\text{ln}\hspace{0.17em}2\right)={\left(e^{\text{ln}\hspace{0.17em}2}\right)}^x\text{ln}\hspace{0.17em}2=2^x\text{ln}\hspace{0.17em}2\text{.}}\\ {\displaystyle \frac{d^{2}y}{d{x}^2}}& =& {\displaystyle \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d{2}^x\text{ln}\hspace{0.17em}2}{dx}=\text{ln}\hspace{0.17em}2·2^x\text{ln}\hspace{0.17em}2={\left(\text{ln}\hspace{0.17em}2\right)}^2{2}^x}\\ {\displaystyle \frac{d^{3}y}{d{x}^3}}& =& {\displaystyle \frac{d\left(\frac{d^{2}y}{d{x}^2}\right)}{dx}=\frac{d}{dx}\left({\left(\text{ln}\hspace{0.17em}2\right)}^2{2}^x\right)={\left(\text{ln}\hspace{0.17em}2\right)}^2{2}^x\text{ln}\hspace{0.17em}2}\\ & =& {\displaystyle {\left(\text{ln}\hspace{0.17em}2\right)}^3{2}^x\text{.}}\end{array}$$

If $y=a\text{cos}\left(\text{ln}\hspace{0.17em}x\right)+b\text{sin}\left(\text{ln}\hspace{0.17em}x\right)$ show that $$x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+y=0\text{.}$$ $$\begin{array}{ccc}{\displaystyle \frac{dy}{dx}}& =& {\displaystyle a(-\text{sin}\left(\text{ln}\hspace{0.17em}x\right))\frac{1}{x}+b\text{cos}\left(\text{ln}\hspace{0.17em}x\right)\frac{1}{x}}\\ & =& {\displaystyle -a\text{sin}\left(\text{ln}\hspace{0.17em}x\right)x^{-1}+b\text{cos}\left(\text{ln}\hspace{0.17em}x\right)x^{-1}\text{.}}\\ {\displaystyle \frac{d^{2}y}{d{x}^2}}& =& {\displaystyle -a\text{cos}\left(\text{ln}\hspace{0.17em}x\right)\frac{1}{x}{x}^{-1}+-a\text{sin}\left(\text{ln}\hspace{0.17em}x\right)(-1)x^{-2}}\\ & =& {\displaystyle +-b\text{sin}\left(\text{ln}\hspace{0.17em}x\right)\frac{1}{x}{x}^{-1}+b\text{cos}\left(\text{ln}\hspace{0.17em}x\right)(-1)x^{-2}}\\ & =& {\displaystyle \frac{-a\text{cos}\left(\text{ln}\hspace{0.17em}x\right)+a\text{sin}\left(\text{ln}\hspace{0.17em}x\right)-b\text{sin}\left(\text{ln}\hspace{0.17em}x\right)-b\text{cos}\left(\text{ln}\hspace{0.17em}x\right)}{x^2}}\\ & =& {\displaystyle \frac{1}{x^2}((a-b)\text{sin}\left(\text{ln}\hspace{0.17em}x\right)-(a+b)\text{cos}\left(\text{ln}\hspace{0.17em}x\right)\text{.})}\end{array}$$ So $$\begin{array}{ccc}{\displaystyle \text{LHS}}& =& {\displaystyle x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+y}\\ & =& {\displaystyle x^2\frac{1}{x}((a-b)\text{sin}\left(\text{ln}\hspace{0.17em}x\right)-(a+b)\text{cos}\left(\text{ln}\hspace{0.17em}x\right))}\\ & & {\displaystyle +x(-a\text{sin}\left(\text{ln}\hspace{0.17em}x\right)x^{-1}+b\text{cos}\left(\text{ln}\hspace{0.17em}x\right)x^{-1})}\\ & & {\displaystyle +a\text{cos}\left(\text{ln}\hspace{0.17em}x\right)+b\text{sin}\left(\text{ln}\hspace{0.17em}x\right)}\\ & =& {\displaystyle (a-b)\text{sin}\left(\text{ln}\hspace{0.17em}x\right)-(a+b)\text{cos}\left(\text{ln}\hspace{0.17em}x\right)}\\ & & {\displaystyle -a\text{sin}\hspace{0.17em}\text{ln}\hspace{0.17em}x+b\text{cos}\hspace{0.17em}\text{ln}\hspace{0.17em}x}\\ & & {\displaystyle +b\text{sin}\hspace{0.17em}\text{ln}\hspace{0.17em}x+a\text{cos}\hspace{0.17em}\text{ln}\hspace{0.17em}x}\\ & =& {\displaystyle 0\text{.}}\end{array}$$

Find $\frac{dy}{dx}$ when $a\text{sin}\left(xy\right)+b\text{cos}\left(\frac{x}{y}\right)=0\text{.}$

Take the derivative: $$\begin{array}{ccc}{\displaystyle 0}& =& {\displaystyle a\text{cos}\left(xy\right)(x\frac{dy}{dx}+1·y)+-b\text{sin}\left(\frac{x}{y}\right)(x(-1)y^{-2}\frac{dy}{dx}+1·y^{-1})}\\ & =& {\displaystyle a\text{cos}\left(xy\right)x\frac{dy}{dx}+a\text{cos}\left(xy\right)y}\\ & & {\displaystyle +b\text{sin}\left(\frac{x}{y}\right)\frac{x}{y^2}\frac{dy}{dx}-b\text{sin}\left(\frac{x}{y}\right)y^{-1}\text{.}}\end{array}$$ Solve for $\frac{dy}{dx}\text{.}$ $$a\text{cos}\left(xy\right)x\frac{dy}{dx}+b\text{sin}\left(\frac{x}{y}\right)\frac{x}{y^2}\frac{dy}{dx}=a\text{cos}\left(xy\right)y-b\text{sin}\left(\frac{x}{y}\right)y^{-1}\text{.}$$ $$\begin{array}{ccc}{\displaystyle \frac{dy}{dx}}& =& {\displaystyle \frac{a\text{cos}\left(xy\right)y-b\text{sin}\left(\frac{x}{y}\right)y^{-1}}{a\text{cos}\left(xy\right)x+b\text{sin}\left(\frac{x}{y}\right)\frac{x}{y^2}}}\\ & =& {\displaystyle \frac{a\text{cos}\left(xy\right)y^2-b\text{sin}\left(\frac{x}{y}\right)}{a\text{cos}\left(xy\right)xy+b\text{sin}\left(\frac{x}{y}\right)\frac{x}{y}}}\\ & =& {\displaystyle \frac{a\text{cos}\left(xy\right)y^3-b\text{sin}\left(\frac{x}{y}\right)}{a\text{cos}\left(xy\right)x{y}^2+b\text{sin}\left(\frac{x}{y}\right)x}\text{.}}\end{array}$$

Find $\frac{dy}{dx}$ when $y={\text{tan}}^{-1}\left(\frac{a}{x}\right)·{\text{cot}}^{-1}\left(\frac{x}{a}\right)\text{.}$ $$\begin{array}{ccc}{\displaystyle \frac{dy}{dx}}& =& {\displaystyle {\text{tan}}^{-1}\left(\frac{a}{x}\right)\left(\frac{-1}{1+{\left(\frac{x}{a}\right)}^2}\right)\frac{1}{a}+\frac{1}{1+{\left(\frac{a}{x}\right)}^2}(-1)a{x}^{-2}{\text{cot}}^{-1}\left(\frac{x}{a}\right)}\\ & =& {\displaystyle \frac{-{\text{tan}}^{-1}\left(\frac{a}{x}\right)}{a+\frac{x^2}{a}}+\frac{-{\text{cot}}^{-1}\left(\frac{x}{a}\right)a}{x^2+a^2}}\\ & =& {\displaystyle \frac{-{\text{tan}}^{-1}\left(\frac{a}{x}\right)a}{a^2+x^2}+\frac{-{\text{cot}}^{-1}\left(\frac{x}{a}\right)a}{x^2+a^2}}\\ & =& {\displaystyle \left(\frac{-a}{a^2+x^2}\right)({\text{tan}}^{-1}\left(\frac{a}{x}\right)+{\text{cot}}^{-1}\left(\frac{x}{a}\right))\text{.}}\end{array}$$ If $\frac{a}{x}=\text{tan}\hspace{0.17em}z$ then $\frac{x}{a}=\text{cot}\hspace{0.17em}z$ and $z={\text{tan}}^{-1}\left(\frac{a}{x}\right)={\text{cot}}^{-1}\left(\frac{x}{a}\right)\text{.}$ So $$\begin{array}{ccc}{\displaystyle \frac{dy}{dx}}& =& {\displaystyle \left(\frac{-a}{a^2+x^2}\right)({\text{tan}}^{-1}\left(\frac{a}{x}\right)+{\text{tan}}^{-1}\left(\frac{a}{x}\right))}\\ & =& {\displaystyle \frac{-2a{\text{tan}}^{-1}\left(\frac{a}{x}\right)}{a^2+x^2}\text{.}}\end{array}$$

Suppose $$f=c_0+c_1(x-a)+c_2{(x-a)}^2+c_3{(x-a)}^3+\cdots \text{.}$$ Find the $c\text{'s.}$ $$\begin{array}{ccc}{\displaystyle f\left(a\right)}& =& {\displaystyle c_0+0+0+0+\cdots =c_0,}\\ {\displaystyle \frac{df}{dx}{|}_{x=a}}& =& {\displaystyle 0+c_1+2c_2(x-a)+3c_3{(x-a)}^2+\cdots {|}_{x=a}}\\ & =& {\displaystyle c_1+0+0+\cdots =c_1,}\\ {\displaystyle \frac{d^{2}f}{d{x}^2}{|}_{x=a}}& =& {\displaystyle 0+2c_2+6c_3(x-a)+4·3{(x-a)}^2+\cdots {|}_{x=a}}\\ & =& {\displaystyle 2c_2+0+0+\cdots =2c_2\text{.}}\end{array}$$ So $$c_2=\frac{1}{2}\left(\frac{d^{2}f}{d{x}^2}{|}_{x=a}\right)\text{.}$$

Find the function $f$ such that $$f\left(10\right)=2,\frac{df}{dx}{|}_{x=10}=-3,\frac{d^{2}f}{d{x}^2}{|}_{x=10}=2,\frac{d^{3}f}{d{x}^3}{|}_{x=10}=1\text{and}\frac{d^{n}f}{d{x}^n}{|}_{x=10}=0$$ for all $n>3\text{.}$

Say $$f=c_0+c_1(x-10)+c_2{(x-10)}^2+c_3{(x-10)}^3+\cdots \text{.}$$ Then $$f\left(10\right)=c_0=2\text{.}$$ So $c_0=2\text{.}$ $$\begin{array}{ccc}{\displaystyle -3=\frac{df}{dx}{|}_{x=10}}& =& {\displaystyle 0+c_1+2c_2(x-10)+3c_3{(x-10)}^2+\cdots {|}_{x=10}}\\ & =& {\displaystyle c_1+0+0+0+\cdots =c_1\text{.}}\end{array}$$ So $c_1=-3\text{.}$ $$\begin{array}{ccc}{\displaystyle 2=\frac{d^{2}f}{d{x}^2}{|}_{x=10}}& =& {\displaystyle 2c_2+3·2c_3(x-10)+4·3{(x-10)}^2+\cdots {|}_{x=10}}\\ & =& {\displaystyle 2c_2+0+0+0+\cdots =2c_2\text{.}}\end{array}$$ So $c_2=1\text{.}$ $$\begin{array}{ccc}{\displaystyle 1=\frac{d^{3}f}{d{x}^3}{|}_{x=10}}& =& {\displaystyle 3·2c_3+4·3·2c_4(x-10)+5·4·3{(x-10)}^2{c}_5+\cdots {|}_{x=10}}\\ & =& {\displaystyle 6c_3+0+0+0+\cdots \text{.}}\end{array}$$ So $c_3=\frac{1}{6}\text{.}$ $$\begin{array}{ccc}{\displaystyle 0=\frac{d^{4}f}{d{x}^4}{|}_{x=10}}& =& {\displaystyle 4·3·2c_4+5·4·3·2c_5(x-10)+\cdots {|}_{x=10}}\\ & =& {\displaystyle 4·3·2c_4+0+0+\cdots =4·3·2c_4\text{.}}\end{array}$$ So $c_4=0\text{.}$ $$\begin{array}{ccc}{\displaystyle 0=\frac{d^{5}f}{d{x}^5}{|}_{x=10}}& =& {\displaystyle 5·4·3·2c_5+6·5·4·3·2(x-10)c_6+\cdots {|}_{x=10}}\\ & =& {\displaystyle 5·4·3·2c_5+0+0+\cdots }\\ & =& {\displaystyle 5!c_5\text{.}}\end{array}$$ So $c_5=0\text{.}$ Similarly $c_6=0,$ $c_7=0,$ $c_8=0,\dots$ since $\frac{d^{n}f}{d{x}^n}{|}_{x=10}=0$ for $n=6,7,8,\dots \text{.}$ So $$\begin{array}{ccc}{\displaystyle f}& =& {\displaystyle c_0+c_1(x-10)+c_2{(x-10)}^2+c_3{(x-10)}^3+\cdots }\\ & =& {\displaystyle 2+-3(x-10)+1·{(x-10)}^2+\frac{1}{6}{(x-10)}^3+0+0+\cdots }\\ & =& {\displaystyle 2-3x+30+x^2-20x+100+\frac{1}{6}(x^3-30x^2+300x-1000)}\\ & =& {\displaystyle \frac{1}{6}{x}^3-4x^2+27x+132-\frac{1000}{6}\text{.}}\end{array}$$ So $$f=\frac{1}{6}{x}^3-4x^2+27x-61\frac{1}{3}\text{.}$$

Notes and References

These are a typed copy of Lecture 9 from a series of handwritten lecture notes for the class MATH 221 given on September 27, 2000.

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