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Q. 86

Expert-verified

Calculus

Found in: Page 199

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

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.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given information

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

Step 2: Find the derivative using z→x

We get,

$\lim_{z \to x} \frac{f (z) - f (x)}{z - x} \lim_{z \to x} \frac{z^{- n} - x^{- n}}{z - x} \lim_{z \to x} \frac{z^{k} - x^{k}}{z - x} [R e p l a c e - n b y k] \lim_{z \to x} \frac{(z - k) (z^{k - 1} + z^{k - 2} x + . . . . + x^{k - 1})}{z - x} \lim_{z \to x} (z^{k - 1} + z^{k - 2} x + . . . . + x^{k - 1}) = x^{k - 1} + x^{k - 1} + . . . . . + x^{k - 1} = k x^{k - 1} N o w r e p l a c e k b y - n = - n x^{- n - 1}$

Step 3: Using h→0

We get,

$\lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \lim_{h \to 0} \frac{{(x + h)}^{- n} - x^{- n}}{h} R e p l a c e - n b y k \lim_{h \to 0} \frac{{(x + h)}^{k} - x^{k}}{h} \lim_{h \to 0} \frac{x^{k} + k x^{k - 1} h + . . . . + h^{k} - x^{k}}{h} \lim_{h \to 0} \frac{k x^{k - 1} h + . . . . + h^{k}}{h} = k x^{k - 1} N o w r e p l a c e k b y - n = - n x^{- n - 1}$

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