MATH 221

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

Last updated: 30 July 2014

Lecture 5

We now know that the derivative with respect to $x$ $$f⟶\begin{array}{c} ddx \end{array}⟶\frac{df}{dx}$$ satisfies

(1)	$\frac{dx}{dx}=1\text{.}$
(2)	$\frac{d(u+v)}{dx}=\frac{du}{dx}+\frac{dv}{dx}\text{.}$
(3)	$\frac{d\left(cf\right)}{dx}=c\frac{df}{dx},$ if $c$ is a constant.
(4)	$\frac{d\left(uv\right)}{dx}=u\frac{db}{dx}+\frac{du}{dx}v\text{.}$
(5)	$\frac{d1}{dx}=0\text{.}$
(6)	$\frac{dx}{dx}=0,$ if $c$ is a constant.
(7)	$\frac{d{x}^n}{dx}=n{x}^{n-1},$ if $n=1,2,3,\dots \text{.}$
(8)	$\frac{d{x}^n}{dx}=n{x}^{n-1},$ if $n=0\text{.}$
(9)	$\frac{d{x}^{-n}}{dx}=(-n)x^{-n-1},$ if $n=1,2,3,\dots \text{.}$
(10)	$\frac{df}{dx}=\frac{df}{dx}\frac{dg}{dx}\text{.}$

Find $\frac{dy}{dx}$ when $y={(2x-5)}^2\text{.}$

If $g=2x-5$ then $y=g^2\text{.}$ So $$\begin{array}{ccc}{\displaystyle \frac{dy}{dx}}& =& {\displaystyle \frac{dy}{dg}\frac{dg}{dx}}\\ & =& {\displaystyle \frac{d{g}^2}{dg}\frac{d(2x-5)}{dx}}\\ & =& {\displaystyle 2g\frac{d(2x-5)}{dx}}\\ & =& {\displaystyle 2(2x-5)·2}\\ & =& {\displaystyle 8x-20\text{.}}\end{array}$$

Find $\frac{dy}{dx}$ when $y={(2x-5)}^2{(3x-4)}^3\text{.}$ $$\begin{array}{ccc}{\displaystyle \frac{dy}{dx}}& =& {\displaystyle \frac{d{(2x-5)}^2{(3x-4)}^3}{dx}}\\ & =& {\displaystyle {(2x-5)}^2\frac{d{(3x-4)}^3}{dx}+\frac{d{(2x-5)}^2}{dx}{(3x-4)}^3}\\ & =& {\displaystyle {(2x-5)}^{2}3{(3x-4)}^2·\frac{d(3x-4)}{dx}+{(3x-4)}^{3}2(2x-5)\frac{d(2x-5)}{dx}}\\ & =& {\displaystyle 3{(2x-5)}^2{(3x-4)}^2·3+2(2x-5){(3x-4)}^3·2}\\ & =& {\displaystyle (2x-5){(3x-4)}^2(9(2x-5)+4(3x-4))}\\ & =& {\displaystyle (2x-5){(3x-4)}^2(30x-61)\text{.}}\end{array}$$

Find $\frac{d{x}^{\frac{m}{n}}}{dx}\text{.}$ $$\begin{array}{ccc}{\displaystyle \frac{d{\left(x^{\frac{m}{n}}\right)}^n}{dx}}& =& {\displaystyle \frac{d{g}^n}{dg}\frac{dg}{dx}\text{if}\hspace{0.17em}g=x^{\frac{m}{n}}}\\ & =& {\displaystyle n{g}^{n-1}\frac{d{x}^{\frac{m}{n}}}{dx}}\\ & =& {\displaystyle n{\left(x^{\frac{m}{n}}\right)}^{n-1}\frac{d{x}^{\frac{m}{n}}}{dx}\text{.}}\end{array}$$ On the other hand $$\frac{d{\left(x^{\frac{m}{n}}\right)}^n}{dx}=\frac{d{x}^m}{dx}=m{x}^{m-1}\text{.}$$ So $$m{x}^{m-1}=n{\left(x\frac{m}{n}\right)}^{n-1}\frac{d{x}^{\frac{m}{n}}}{dx}\text{.}$$ So $$\frac{d{x}^{\frac{m}{n}}}{dx}=\frac{m{x}^{m-1}}{n{\left(x^{\frac{m}{n}}\right)}^n{\left(x^{\frac{m}{n}}\right)}^{-1}}\text{.}$$ So $$\frac{d{x}^{\frac{m}{n}}}{dx}=\frac{m{x}^{m-1}{x}^{\frac{m}{n}}}{n{x}^m}=\frac{m}{n}{x}^{\frac{m}{n}}{x}^{-1}\text{.}$$ So $$\frac{d{x}^{\frac{m}{n}}}{dx}=\frac{m}{n}{x}^{\frac{m}{n}-1}\text{.}$$

Find $\frac{dy}{dx}$ if $y=\frac{\sqrt{1+x^2}}{\sqrt{1-x^2}}\text{.}$ $$\begin{array}{ccc}{\displaystyle \frac{dy}{dx}}& =& {\displaystyle \frac{d\frac{\sqrt{1+x^2}}{\sqrt{1-x^2}}}{dx}}\\ & =& {\displaystyle \frac{d\frac{{(1+x^2)}^{\frac{1}{2}}}{{(1-x^2)}^{\frac{1}{2}}}}{dx}}\\ & =& {\displaystyle \frac{d{\left(\frac{1+x^2}{1-x^2}\right)}^{\frac{1}{2}}}{dx}}\\ & =& {\displaystyle \frac{1}{2}{\left(\frac{1+x^2}{1-x^2}\right)}^{\frac{1}{2}-1}\frac{d\left(\frac{1+x^2}{1-x^2}\right)}{dx}=\frac{1}{2}{\left(\frac{1+x^2}{1-x^2}\right)}^{-\frac{1}{2}}\frac{d(1+x^2){(1-x^2)}^{-1}}{dx}}\\ & =& {\displaystyle \frac{1}{2}{\left(\frac{1+x^2}{1-x^2}\right)}^{\frac{1}{2}}((1+x^2)\frac{d{(1-x^2)}^{-1}}{dx}+\frac{d(1+x^2)}{dx}{(1-x^2)}^{-1})}\\ & =& {\displaystyle \frac{1}{2}{\left(\frac{1+x^2}{1-x^2}\right)}^{\frac{1}{2}}((1+x^2)(-1){(1-x^2)}^{-2}\frac{d(1-x^2)}{dx}+2x{(1-x^2)}^{-1})}\\ & =& {\displaystyle \frac{1}{2}{\left(\frac{1+x^2}{1-x^2}\right)}^{\frac{1}{2}}(\frac{(-1)(1+x^2)(-2x)}{{(1-x^2)}^2}+\frac{2x}{1-x^2})}\\ & =& {\displaystyle \frac{1}{2}{\left(\frac{1+x^2}{1-x^2}\right)}^{\frac{1}{2}}(\frac{2x(1+x^2)}{{(1-x^2)}^2}+\frac{2x(1-x^2)}{{\left(1-x^2\right)}^2})}\\ & =& {\displaystyle \frac{1}{2}{\left(\frac{1+x^2}{1-x^2}\right)}^{\frac{1}{2}}\left(\frac{2x(1+x^2+1-x^2)}{{(1-x^2)}^2}\right)}\\ & =& {\displaystyle \frac{1}{2}\frac{{(1-x^2)}^{\frac{1}{2}}}{{(1+x^2)}^{\frac{1}{2}}}\frac{4x}{{(1-x^2)}^2}}\\ & =& {\displaystyle \frac{2x}{{(1+x^2)}^{\frac{1}{2}}{(1-x^2)}^{\frac{3}{2}}}\text{.}}\end{array}$$

Differentiate $\frac{x^2}{1+x^2}$ with respect to $x^2\text{.}$

This is the same problem as: Find $\frac{dz}{dp}$ when $z=\frac{x^2}{1+x^2}$ and $p=x^2\text{.}$ Now $$\frac{dz}{dx}=\frac{dz}{dp}\frac{dp}{dx}\text{.}$$ So $$\frac{dz}{dp}=\frac{\frac{dz}{dx}}{\frac{dp}{dx}}\text{.}$$ So $$\begin{array}{ccc}{\displaystyle \frac{dz}{dp}}& =& {\displaystyle \frac{\frac{d\left(\frac{x^2}{1+x^2}\right)}{dx}}{\frac{d\left(x^2\right)}{dx}}}\\ & =& {\displaystyle \frac{\frac{d{x}^2{(1+x^2)}^{-1}}{dx}}{\frac{d{x}^2}{dx}}}\\ & =& {\displaystyle \frac{\frac{d{x}^2{(1+x^2)}^{-1}}{dx}}{2x}}\\ & =& {\displaystyle \frac{x^2\frac{d{(1+x^2)}^{-1}}{dx}+\frac{d{x}^2}{dx}{(1+x^2)}^{-1}}{2x}}\\ & =& {\displaystyle \frac{x^2(-1){(1+x^2)}^{-2}\frac{d(1+x^2)}{dx}+2x{(1+x^2)}^{-1}}{2x}}\\ & =& {\displaystyle \frac{\frac{-x^2}{{(1+x^2)}^2}2x+\frac{2x}{1+x^2}}{2x}}\\ & =& {\displaystyle \frac{-x^2}{{(1+x^2)}^2}+\frac{1}{1+x^2}}\\ & =& {\displaystyle \frac{-x^2+1+x^2}{{(1+x^2)}^2}}\\ & =& {\displaystyle \frac{1}{{(1+x^2)}^2}\text{.}}\end{array}$$

Find $\frac{dy}{dx}$ when $x^4+y^4=4a^2{x}^2{y}^2\text{.}$ $$\frac{d(x^4+y^4)}{dx}=\frac{d4a^2{x}^2{y}^2}{dx}\text{.}$$ So $$\frac{d{x}^4}{dx}+\frac{d{y}^4}{dx}=4a^2\frac{d{x}^2{y}^2}{dx}\text{.}$$ So $$4x^3+4y^3\frac{dy}{dx}=4a^2(x^2\frac{d{y}^2}{dx}+\frac{d{x}^2}{dx}{y}^2)\text{.}$$ So $$\begin{array}{ccc}{\displaystyle 4x^3+4y^3\frac{dy}{dx}}& =& {\displaystyle 4a^2(x^{2}2y\frac{dy}{dx}+2x{y}^2)}\\ & =& {\displaystyle 4a^2{x}^{2}2y\frac{dy}{dx}+4a^{2}2x{y}^2\text{.}}\end{array}$$ So $$4x^3-4a^{2}2x{y}^2=4a^2{x}^{2}2y\frac{dy}{dx}-4y^3\frac{dy}{dx}\text{.}$$ So $$4x^3-4a^{2}2x{y}^2=(4a^2{x}^{2}2y-4y^3)\frac{dy}{dx}\text{.}$$ So $$\frac{4x^3-4a^{2}2x{y}^2}{4a^2{x}^{2}2y-4y^3}=\frac{dy}{dx}\text{.}$$ So $$\frac{dy}{dx}=\frac{x^3-2a^{2}x{y}^2}{2a^2{x}^{2}y-y^3}\text{.}$$ All we did is take the derivative of both sides and then solve for $\frac{d}{dx}\text{.}$

Find $\frac{dy}{dx}{|}_{x=3}$ when $y=(x+1)(x+2)\text{.}$

$\frac{dy}{dx}{|}_{x=3}$ means: find $\frac{dy}{dx}$ and then plug in $x=3\text{.}$ $$\begin{array}{ccc}{\displaystyle $ \frac{dy}{dx}{|}_{x=3}$}& =& {\displaystyle \frac{d(x+1)(x+2)}{dx}{|}_{x=3}}\\ & =& {\displaystyle ((x+1)\frac{d(x+2)}{dx}+\frac{d(x+1)}{dx}(x+2)){|}_{x=3}}\\ & =& {\displaystyle ((x+1)+(x+2)){|}_{x=3}=(2x+3){|}_{x=3}}\\ & =& {\displaystyle 2·3+3=9\text{.}}\end{array}$$

Find $\frac{dy}{dx}$ when $x=\frac{3at}{1+t^3}$ and $y=\frac{3a{t}^2}{1+t^3}\text{.}$ $$\frac{dy}{dx}=\frac{d\left(xt\right)}{dx}\text{since}y=\frac{3a{t}^2}{1+t^3}=\left(\frac{3at}{1+t^3}\right)t=xt\text{.}$$ So $$\frac{dy}{dx}=x\frac{dt}{dx}+\frac{dx}{dx}t=x\frac{dt}{dx}+t\text{.}$$ What is $\frac{dt}{dx}\text{??}$ Since $\frac{dx}{dx}=\frac{dx}{dt}\frac{dt}{dx},$ $\frac{dt}{dx}=\frac{\frac{dx}{dx}}{\frac{dx}{dt}}=\frac{1}{\frac{dx}{dt}}\text{.}$ So $$\begin{array}{ccc}{\displaystyle \frac{dt}{dx}}& =& {\displaystyle \frac{1}{\frac{dx}{dt}}=\frac{1}{\frac{d\left(\frac{3at}{1+t^3}\right)}{dt}}=\frac{1}{\frac{d\left(3at\right){(1+t^3)}^{-1}}{dt}}}\\ & =& {\displaystyle \frac{1}{3at(-1){(1+t^3)}^{-2}\frac{d(1+t^3)}{dt}+3a{(1+t^3)}^{-1}}}\\ & =& {\displaystyle \frac{1}{\frac{-3at}{{(1+t^3)}^2}·3t^2+\frac{3a}{1+t^3}}}\\ & =& {\displaystyle \frac{1}{\frac{-9a{t}^3+3a(1+t^3)}{{(1+t^3)}^2}}}\\ & =& {\displaystyle \frac{{(1+t^3)}^2}{-9a{t}^3+3a{t}^3+3a}}\\ & =& {\displaystyle \frac{{(1+t^3)}^2}{3a-6a{t}^3}\text{.}}\end{array}$$ So $$\begin{array}{ccc}{\displaystyle \frac{dy}{dx}}& =& {\displaystyle x\frac{dt}{dx}+t}\\ & =& {\displaystyle \frac{3at}{1+t^3}\frac{{(1+t^3)}^2}{3a(1-2t^3)}+t}\\ & =& {\displaystyle \frac{t(1+t^3)}{1-2t^3}+\frac{t(1-2t^3)}{1-2t^3}}\\ & =& {\displaystyle \frac{t+t^4+t-2t^4}{1-2t^3}}\\ & =& {\displaystyle \frac{2t-t^4}{1-2t^3}\text{.}}\end{array}$$

Notes and References

These are a typed copy of Lecture 5 from a series of handwritten lecture notes for the class MATH 221 given on September 15, 2000.

page history