Table of Contents

Problem 3

Author: Philip Nelson

a.

\[\frac{d}{dx} \sqrt{x}\]

\[=\frac{\sqrt(x+h)-\sqrt(x)}{h}\]

\[=\frac{\sqrt(x+h)-\sqrt(x)}{h} \cdot \frac{\sqrt(x+h)+\sqrt(x)}{\sqrt(x+h)+\sqrt(x)}\]

\[=\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}\]

\[=\frac{h}{h(\sqrt{x+h}+\sqrt{x})}\]

\[=\frac{1}{\sqrt{x+h}+\sqrt{x}}\]

Original Approximation \[\frac{d}{dx} \sqrt{x} = \frac{\sqrt(x+h)-\sqrt(x)}{h}\]

New Approximation \[\frac{d}{dx} \sqrt{x} = \frac{1}{\sqrt{x+h}+\sqrt{x}}\]

b. Using the difference \[\frac{d}{dx} f(x) = \frac{f(x+h)-f(x)}{h}\] where \(f(x) = \sin(x)\) at \(x_0 = 0\) and using the Taylor series expansion for \(\sin(x)\) \[\sum_{k=0}^5 \frac{(-1)^k}{(2k+1)!}x^{2k+1}\] the following graphs were produced to visualize the value of the derivative and the associated error.

Last Modified: September 2018