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Q. 86

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Found in: Page 199

### Calculus

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 powersa) by using the $z\to x$ definition of the derivativeb) by using the $h\to 0$ definition of the derivative

We prove the power rule for negative powers.

See the step by step solution

## Step 1: Given information

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

## Step 2: Find the derivative using z→x

We get,

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

## Step 3: Using h→0

We get,

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