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Q. 86

Expert-verifiedFound in: Page 199

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Prove, in two ways, that the power rule holds for negative integer powers

a) by using the $z\to x$ definition of the derivative

b) by using the $h\to 0$ definition of the derivative

We prove the power rule for negative powers.

We are given a function $f\left(x\right)={x}^{-n}$

We get,

$\underset{z\to x}{\mathrm{lim}}\frac{f\left(z\right)-f\left(x\right)}{z-x}\phantom{\rule{0ex}{0ex}}\underset{z\to x}{\mathrm{lim}}\frac{{z}^{-n}-{x}^{-n}}{z-x}\phantom{\rule{0ex}{0ex}}\underset{z\to x}{\mathrm{lim}}\frac{{z}^{k}-{x}^{k}}{z-x}[Replace-nbyk]\phantom{\rule{0ex}{0ex}}\underset{z\to x}{\mathrm{lim}}\frac{(z-k)({z}^{k-1}+{z}^{k-2}x+....+{x}^{k-1})}{z-x}\phantom{\rule{0ex}{0ex}}\underset{z\to x}{\mathrm{lim}}({z}^{k-1}+{z}^{k-2}x+....+{x}^{k-1})\phantom{\rule{0ex}{0ex}}={x}^{k-1}+{x}^{k-1}+.....+{x}^{k-1}\phantom{\rule{0ex}{0ex}}=k{x}^{k-1}\phantom{\rule{0ex}{0ex}}Nowreplacekby-n\phantom{\rule{0ex}{0ex}}=-n{x}^{-n-1}$

We get,

$\underset{h\to 0}{\mathrm{lim}}\frac{f(x+h)-f\left(x\right)}{h}\phantom{\rule{0ex}{0ex}}\underset{h\to 0}{\mathrm{lim}}\frac{{(x+h)}^{-n}-{x}^{-n}}{h}\phantom{\rule{0ex}{0ex}}Replace-nbyk\phantom{\rule{0ex}{0ex}}\underset{h\to 0}{\mathrm{lim}}\frac{{(x+h)}^{k}-{x}^{k}}{h}\phantom{\rule{0ex}{0ex}}\underset{h\to 0}{\mathrm{lim}}\frac{{x}^{k}+k{x}^{k-1}h+....+{h}^{k}-{x}^{k}}{h}\phantom{\rule{0ex}{0ex}}\underset{h\to 0}{\mathrm{lim}}\frac{k{x}^{k-1}h+....+{h}^{k}}{h}\phantom{\rule{0ex}{0ex}}=k{x}^{k-1}\phantom{\rule{0ex}{0ex}}Nowreplacekby-n\phantom{\rule{0ex}{0ex}}=-n{x}^{-n-1}$

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