Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Appendices - Section A.2 - Mathematical Induction - Exercises A.2 - Page AP-9: 3


See proof below.

Work Step by Step

$1. \ (\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e})$ Check that the formula holds for n = 1. $LHS= \displaystyle \frac{d}{dx}[x^{1}]=1\qquad $ (we are given this by the problem text) $RHS=1\cdot x^{1-1}=1\cdot x^{0}=1\cdot 1=1=LHS$ $\qquad $ yes, it is true when $n=1.$ $2. \ (\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{p})$ Prove that IF the formula holds for any positive integer $n=k$, THEN it also holds for the next integer, $n=k+1$. Let us assume that for $n=k$, $ \displaystyle \frac{d}{dx}[x^{k}]=kx^{k-1}$ (we call this the step hypothesis) For $n=k+1$, the LHS equals $ \displaystyle \frac{d}{dx}[x^{k+1}]$, which we rewrite as $LHS= \displaystyle \frac{d}{dx}[x^{k}\cdot x^{1}]\qquad $ Apply the product rule with $\left[\begin{array}{l} u=x^{k}\\ v=x \end{array}\right]$ $=(\displaystyle \frac{d}{dx}u)\cdot v+u(\frac{d}{dx}v)$ $=\displaystyle \frac{d}{dx}[x^{k}]\cdot x+x^{k}(\frac{d}{dx}[x])$ The first term contains $\displaystyle \frac{d}{dx}[x^{k}] $ which, by the hypothesis, equals $kx^{k-1}$ In the second term, we are given by the text: $\displaystyle \frac{d}{dx}[x]=1.$ $=kx^{k-1}\cdot x+x^{k}\cdot 1$ $=kx^{k}+x^{k}\qquad$ ... factor out $x^{k}$ $=x^{k}(k+1)$ $=(k+1)\cdot x^{(k+1)-1}=RHS$ for $n=k+1$ which makes the formula true for $n=k+1.$ By mathematical induction, the formula is true for any integer $n.$
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