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 10

Finding derivatives with limits

If $f$ is a function then $$f\left(x\right)=f\left(a\right)+\left(\frac{df}{dx}{|}_{x=a}\right)(x-a)+\frac{1}{2!}\left(\frac{d^{2}f}{d{x}^2}{|}_{x=a}\right){(x-a)}^2+\frac{1}{3!}\left(\frac{d^{3}f}{d{x}^3}{|}_{x=a}\right){(x-a)}^3+\cdots \text{.}$$ Substitute $x=a+\mathrm{\Delta }x\text{.}$ $\mathrm{\Delta }x$ stands for a small change in $x\text{.}$ $$f(a+\mathrm{\Delta }x)=f\left(a\right)+\left(\frac{df}{dx}{|}_{x=a}\right)\left(\mathrm{\Delta }x\right)+\frac{1}{2!}\left(\frac{d^{2}f}{d{x}^2}{|}_{x=a}\right){\left(\mathrm{\Delta }x\right)}^2+\frac{1}{3!}\left(\frac{d^{3}f}{d{x}^3}{|}_{x=a}\right){\left(\mathrm{\Delta }x\right)}^3+\cdots \text{.}$$ So $$f(a+\mathrm{\Delta }x)-f\left(a\right)=\left(\frac{df}{dx}{|}_{x=a}\right)\mathrm{\Delta }x+\frac{1}{2!}\left(\frac{d^{2}f}{d{x}^2}{|}_{x=a}\right){\left(\mathrm{\Delta }x\right)}^2+\cdots \text{.}$$ So $$\frac{f(a+\mathrm{\Delta }x)-f\left(a\right)}{\mathrm{\Delta }x}=\frac{df}{dx}{|}_{x=a}+\frac{1}{2!}\left(\frac{d^{2}d}{d{x}^2}{|}_{x=a}\right)\mathrm{\Delta }x+\frac{1}{3!}\left(\frac{d^{3}f}{d{x}^3}{|}_{x=a}\right){\left(\mathrm{\Delta }x\right)}^2+\cdots \text{.}$$ So $$\underset{\mathrm{\Delta }x\to 0}{\text{lim}}\frac{f(a+\mathrm{\Delta }x)-f\left(a\right)}{\mathrm{\Delta }x}=\frac{df}{dx}{|}_{x=a}\text{.}$$

Suppose $f\left(x\right)=x^3\text{.}$ What is $f\left(3.02\right)\text{?}$ $$\begin{array}{ccc}{\displaystyle f\left(3.02\right)}& =& {\displaystyle f(3+.02)}\\ & =& {\displaystyle f\left(3\right)+\left(\frac{df}{dx}\right)\left(.02\right)+\frac{1}{2}\left(\frac{d^{2}f}{d{x}^2}\right){\left(.02\right)}^2+\cdots }\end{array}$$ since $$f(a+\mathrm{\Delta }x)=f\left(a\right)+\left(\frac{df}{dx}{|}_{x=a}\right)\mathrm{\Delta }x+\frac{1}{2}\left(\frac{d^{2}f}{d{x}^2}{|}_{x=a}\right){\left(\mathrm{\Delta }x\right)}^2+\cdots \text{.}$$ Now $$\begin{array}{c}{\displaystyle f\left(3\right)=27,\frac{df}{dx}{|}_{x=3}=3x^2{|}_{x=3}=27,}\\ {\displaystyle \frac{d^{2}f}{d{x}^2}{|}_{x=3}=6x{|}_{x=3}=18,\frac{d^{3}f}{d{x}^3}{|}_{x=3}=6{|}_{x=3}=6,}\\ {\displaystyle \frac{d^{4}f}{d{x}^4}{|}_{x=3}=0{|}_{x=3}=0,\frac{d^{5}f}{d{x}^5}{|}_{x=3}=0{|}_{x=3}=0,\dots }\end{array}$$ So $$\begin{array}{ccc}{\displaystyle f\left(3.02\right)}& =& {\displaystyle f(3+.02)}\\ & =& {\displaystyle 27+27\left(.02\right)+\frac{1}{2}18{\left(.02\right)}^2+\frac{6}{3!}{\left(.02\right)}^3+0+0+\cdots }\\ & =& {\displaystyle 27+.54+9\left(.0004\right)+.000008}\\ & =& {\displaystyle 27.543608\text{.}}\end{array}$$

What is the expansion of $f(a+\mathrm{\Delta }x)$ when $f\left(x\right)=e^{3x}$ and $a=0\text{?}$ $$\begin{array}{ccc}{\displaystyle f(a+\mathrm{\Delta }x)}& =& {\displaystyle e^{3(a+\mathrm{\Delta }x)}=e^{3\mathrm{\Delta }x}}\\ & =& {\displaystyle 1+3\mathrm{\Delta }x+\frac{{\left(3\mathrm{\Delta }x\right)}^2}{2!}+\frac{{\left(3\mathrm{\Delta }x\right)}^3}{3!}+\frac{{\left(3\mathrm{\Delta }x\right)}^4}{4!}+\cdots }\\ & =& {\displaystyle 1+3\mathrm{\Delta }x+\frac{9}{2!}{\left(\mathrm{\Delta }x\right)}^2+\frac{27}{3!}{\left(\mathrm{\Delta }x\right)}^3+\frac{3^4{\left(\mathrm{\Delta }x\right)}^4}{4!}+\cdots \text{.}}\end{array}$$ Second way: $$\begin{array}{ccc}{\displaystyle f(a+\mathrm{\Delta }x)}& =& {\displaystyle f(0+\mathrm{\Delta }x)}\\ & =& {\displaystyle f\left(0\right)+\frac{df}{dx}{|}_{x=0}\mathrm{\Delta }x+\frac{1}{2}\left(\frac{d^{2}f}{d{x}^2}{|}_{x=0}\right){\left(\mathrm{\Delta }x\right)}^2+\cdots \text{.}}\end{array}$$ $$\begin{array}{c}{\displaystyle f\left(0\right)=e^{3·0}=1,\frac{df}{dx}{|}_{x=0}=\frac{d{e}^{3x}}{dx}{|}_{x=0}=3e^{3x}{|}_{x=0}=3,}\\ {\displaystyle \frac{d^{2}f}{d{x}^2}{|}_{x=0}=\frac{d3e^{3x}}{dx}{|}_{x=0}=3^2{e}^{3x}{|}_{x=0}=3^2,}\\ {\displaystyle \frac{d^{3}f}{d{x}^3}{|}_{x=0}=\frac{d{3}^2{e}^{3x}}{dx}{|}_{x=0}=3^3{e}^{3x}{|}_{x=0}=3^3,\dots \text{.}}\end{array}$$ So $$f(0+\mathrm{\Delta }x)=e^{3\mathrm{\Delta }x}=1+3\mathrm{\Delta }x+\frac{1}{2}{3}^2{\left(\mathrm{\Delta }x\right)}^2+\frac{1}{3!}{3}^3{\left(\mathrm{\Delta }x\right)}^3+\cdots \text{.}$$

If $f\left(x\right)=\text{ln}(1+x)$ expand $f(a+\mathrm{\Delta }x)$ in terms of $\mathrm{\Delta }x$ when $a=0\text{.}$ $$\begin{array}{ccc}{\displaystyle f(a+\mathrm{\Delta }x)}& =& {\displaystyle f(0+\mathrm{\Delta }x)=f\left(\mathrm{\Delta }x\right)=\text{ln}(1+\mathrm{\Delta }x)}\\ & =& {\displaystyle f\left(0\right)+\left(\frac{df}{dx}{|}_{x=0}\right)\mathrm{\Delta }x+\frac{1}{2}\left(\frac{d^{2}f}{d{x}^2}{|}_{x=0}\right){\left(\mathrm{\Delta }x\right)}^2+\frac{1}{3!}\left(\frac{d^{3}f}{d{x}^3}{|}_{x=0}\right){\left(\mathrm{\Delta }x\right)}^3+\cdots }\\ & =& {\displaystyle \text{ln}(1+0)=\frac{d\text{ln}(1+x)}{dx}{|}_{x=0}+\frac{1}{2!}\left(\frac{d^2\text{ln}(1+x)}{d{x}^2}{|}_{x=0}\right){\left(\right)}^2+\cdots ,}\\ {\displaystyle \text{ln}(1+0)}& =& {\displaystyle \text{ln}\hspace{0.17em}1=0,}\\ {\displaystyle \frac{d\text{ln}(1+x)}{dx}{|}_{x=0}}& =& {\displaystyle \frac{1}{1+x}{|}_{x=0}=\frac{1}{1}=1,}\\ {\displaystyle \frac{d^2\text{ln}(1+x)}{d{x}^2}{|}_{x=0}}& =& {\displaystyle \frac{d\frac{1}{1+x}}{dx}{|}_{x=0}=\frac{-1}{{(1+x)}^2}{|}_{x=0}=-1,}\\ {\displaystyle \frac{d^3\text{ln}(1+x)}{d{x}^3}{|}_{x=0}}& =& {\displaystyle \frac{d\frac{-1}{{(1+x)}^2}}{dx}{|}_{x=0}=\frac{(-1)(-2)}{{(1+x)}^3}{|}_{x=0}=2·1,}\\ {\displaystyle \frac{d^4\text{ln}(1+x)}{d{x}^4}{|}_{x=0}}& =& {\displaystyle \frac{d\frac{2!}{{(1+x)}^3}}{dx}{|}_{x=0}=\frac{-3·2·1}{{(1+x)}^4}{|}_{x=0}=-3!\text{.}}\end{array}$$ So $$\begin{array}{ccc}{\displaystyle \text{ln}(1+\mathrm{\Delta }x)}& =& {\displaystyle 0+(+1)\mathrm{\Delta }x+\frac{1}{2!}(-1){\left(\mathrm{\Delta }x\right)}^2+\frac{1}{3!}2!{\left(\mathrm{\Delta }x\right)}^3+\frac{1}{4!}(-3!){\left(\mathrm{\Delta }x\right)}^4+\cdots }\\ & =& {\displaystyle 0+\mathrm{\Delta }x-\frac{{\left(\mathrm{\Delta }x\right)}^2}{2}+\frac{{\left(\mathrm{\Delta }x\right)}^3}{3}-\frac{{\left(\mathrm{\Delta }x\right)}^4}{4}+\cdots }\\ & =& {\displaystyle \mathrm{\Delta }x-\frac{{\left(\mathrm{\Delta }x\right)}^2}{2}+\frac{{\left(\mathrm{\Delta }x\right)}^3}{3}-\frac{{\left(\mathrm{\Delta }x\right)}^4}{4}+\cdots \text{.}}\end{array}$$ The linear approximation to $\text{ln}(1+x)$ at $x=0$ is $$\text{ln}(1+\mathrm{\Delta }x)\approx +\mathrm{\Delta }x\text{.}$$ The quadratic approximation to $\text{ln}(1+x)$ at $x=0$ is $$\text{ln}(1+\mathrm{\Delta }x)\approx \mathrm{\Delta }x-\frac{{\left(\mathrm{\Delta }x\right)}^2}{2!}\text{.}$$

Approximate the value of ${\left(255\right)}^{\frac{1}{4}}\text{.}$

Let $f\left(x\right)=x^{\frac{1}{4}}\text{.}$ Then $${\left(255\right)}^{\frac{1}{4}}={(256-1)}^{\frac{1}{4}}=f(a+\mathrm{\Delta }x)$$ with $a=256$ and $\mathrm{\Delta }x=-1\text{.}$ This is an approximation to ${255}^{\frac{1}{4}}$ using a linear approximation. $$\begin{array}{ccc}{\displaystyle f(a+\mathrm{\Delta }x)}& \approx & {\displaystyle f\left(a\right)+\left(\frac{df}{dx}{|}_{x=a}\right)\mathrm{\Delta }x+\frac{1}{2}\left(\frac{d^{2}f}{d{x}^2}{|}_{x=a}\right){\left(\mathrm{\Delta }x\right)}^2}\\ & =& {\displaystyle a^{\frac{1}{4}}+\left(\frac{1}{4}{x}^{-\frac{3}{4}}{|}_{x=a}\right)\mathrm{\Delta }x+\frac{1}{2}\left(\frac{1}{4}\right)\left(-\frac{3}{4}\right)x^{-\frac{7}{4}}{|}_{x=a}{\left(\mathrm{\Delta }x\right)}^2}\\ & =& {\displaystyle a^{\frac{1}{4}}+\frac{1}{4}{\left(a^{\frac{1}{4}}\right)}^{-3}\mathrm{\Delta }x+-\frac{3}{2}\frac{1}{4^2}{\left(a^{\frac{1}{4}}\right)}^{-7}{\left(\mathrm{\Delta }x\right)}^2}\\ & =& {\displaystyle 4+\frac{1}{4}{4}^{-3}(-1)-\frac{3}{2·4^2}{4}^{-7}{(-1)}^2}\\ & =& {\displaystyle 4-\frac{1}{4^4}+\frac{3}{2·4^9}=4-\frac{1}{16}+\frac{3}{2.262144}}\\ & =& {\displaystyle 3.99609947204589843750\text{.}}\end{array}$$ This is the quadratic approximation to ${255}^{\frac{1}{4}}\text{.}$ The correct answer is $${255}^{\frac{1}{4}}=3.9960880148804670689\dots$$ according to my computer. The computer is clearly wrong (it is off by at least $.000005\text{).}$

Expand $f(a+\mathrm{\Delta }x)$ in terms of $\mathrm{\Delta }x$ when $a=4$ and $f\left(x\right)=\sqrt{x}\text{.}$ $$\begin{array}{ccc}{\displaystyle f(a+\mathrm{\Delta }x)}& =& {\displaystyle f(4+\mathrm{\Delta }x)=\sqrt{4+\mathrm{\Delta }x}}\\ & =& {\displaystyle f\left(4\right)+\left(\frac{df}{dx}{|}_{x=4}\right)\mathrm{\Delta }x+\frac{1}{2!}\left(\frac{d^{2}f}{d{x}^2}{|}_{x=4}\right){\left(\mathrm{\Delta }x\right)}^2+\cdots ,}\\ {\displaystyle f\left(4\right)}& =& {\displaystyle \sqrt{4}=2\text{.}}\\ {\displaystyle \frac{df}{dx}{|}_{x=4}}& =& {\displaystyle \frac{d{x}^{\frac{1}{2}}}{dx}{|}_{x=4}=\frac{1}{2}{x}^{-\frac{1}{2}}{|}_{x=4}=\frac{1}{2}·\frac{1}{2}=\frac{1}{4},}\\ {\displaystyle \frac{d^{2}f}{d{x}^2}{|}_{x=4}}& =& {\displaystyle \frac{d\frac{1}{2}{x}^{-\frac{1}{2}}}{dx}{|}_{x=4}=\frac{1}{2}(-\frac{1}{2})x^{-\frac{3}{2}}{|}_{x=4}=\frac{-1}{2·2}\frac{1}{2^3}=\frac{-1}{2^5},}\\ {\displaystyle \frac{d^{3}f}{d{x}^3}{|}_{x=4}}& =& {\displaystyle \frac{d\frac{-1}{2^2}{x}^{-\frac{3}{2}}}{dx}{|}_{x=4}=\frac{-1}{2^2}\frac{-3}{2}{x}^{-\frac{5}{2}}{|}_{x=4}=\frac{3}{2^3}\frac{1}{2^5}=\frac{3}{2^8},}\\ {\displaystyle \frac{d^{4}f}{d{x}^4}{|}_{x=4}}& =& {\displaystyle \frac{d\frac{3}{2^3}{x}^{-\frac{5}{2}}}{dx}{|}_{x=4}=\frac{3}{2^3}(-\frac{5}{2})x^{-\frac{7}{2}}{|}_{x=4}=\frac{-3·5}{2^4}\frac{1}{2^7}=\frac{-3·5}{2^{11}},}\\ {\displaystyle \cdots }& & {\displaystyle }\end{array}$$ So $$\begin{array}{ccc}{\displaystyle \sqrt{4+\mathrm{\Delta }x}}& =& {\displaystyle f(4+\mathrm{\Delta }x)}\\ & =& {\displaystyle 2+\frac{1}{2^2}\mathrm{\Delta }x-\frac{1}{2!}\frac{1}{2^5}{\left(\mathrm{\Delta }x\right)}^2+\frac{1}{3!}\frac{3}{2^8}{\left(\mathrm{\Delta }x\right)}^3-\frac{1}{4!}\frac{3·5}{2^{11}}{\left(\mathrm{\Delta }x\right)}^4+\cdots \text{.}}\end{array}$$ The linear approximation to $\sqrt{x}$ at $x=4$ is $$\sqrt{4+\mathrm{\Delta }x}\approx 2+\frac{1}{4}\mathrm{\Delta }x\text{.}$$ The quadratic approximation to $\sqrt{x}$ at $x=4$ is $$\sqrt{4+\mathrm{\Delta }x}\approx 2+\frac{1}{4}\mathrm{\Delta }x-\frac{1}{64}{\left(\mathrm{\Delta }x\right)}^2\text{.}$$

Approximate $\sqrt{4.03}$.

Linear: $$\sqrt{4.03}\approx 2+\frac{1}{4}\left(.03\right)=2.0075\text{.}$$ Quadratic: $$\sqrt{4.03}\approx 2+\frac{1}{4}\left(.03\right)-\frac{1}{64}{\left(.03\right)}^2=2.0074859375\text{.}$$

Notes and References

These are a typed copy of Lecture 10 from a series of handwritten lecture notes for the class MATH 221 given on September 29, 2000.

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