Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 3 - Review - Concept Check - Page 269: 2

Answer

(a) $y=x^{n} \Rightarrow y^{\prime}=n x^{n-1}$ (b) $y=e^{x} \Rightarrow y^{\prime}=e^{x}$ (c) $y=b^{x} \Rightarrow y^{\prime}=b^{x} \ln b$ (d) $y=\ln x \Rightarrow y^{\prime}=1 / x$ (e) $y=\log _{b} x \Rightarrow y^{\prime}=1 /(x \ln b)$ (f) $y=\sin x \Rightarrow y^{\prime}=\cos x$ (g) $y=\cos x \Rightarrow y^{\prime}=-\sin x$ (h) $y=\tan x \Rightarrow y=\sec ^{2} x$ (i) $y=\csc x \Rightarrow y^{\prime}=-\operatorname{esc} x \cot x$ (j) $y=\sec x \Rightarrow y^{\prime}=\sec x \tan x$ (k) $y=\cot x \Rightarrow y^{\prime}=-\csc ^{2} x$ (l) $y=\sin ^{-1} x=y^{\prime}=1 / \sqrt{1-x^{2}}$ (m) $y=\cos ^{-1} x \Rightarrow y^{\prime}=-1 / \sqrt{1-x^{2}}$ (n) $y=\tan ^{-1} x \Rightarrow \quad y^{\prime}=1 /\left(1+x^{2}\right)$ (0) $y=\sinh x \Rightarrow y^{\prime}=\cosh x$ (p) $y=\cosh x \Rightarrow y^{\prime}=\sinh x$ (q) $y=\tanh x \Rightarrow y^{\prime}=\operatorname{sech}^{2} x$ (r) $y=\sinh ^{-1} x \Rightarrow y^{\prime}=1 / \sqrt{1+x^{2}}$ (5) $y=\cosh ^{-1} x \Rightarrow y^{\prime}=1 / \sqrt{x^{2}-1}$ (t) $y=\tanh ^{-1} x \Rightarrow y^{\prime}=1 /\left(1-x^{2}\right)$

Work Step by Step

(a) $y=x^{n} \Rightarrow y^{\prime}=n x^{n-1}$ (b) $y=e^{x} \Rightarrow y^{\prime}=e^{x}$ (c) $y=b^{x} \Rightarrow y^{\prime}=b^{x} \ln b$ (d) $y=\ln x \Rightarrow y^{\prime}=1 / x$ (e) $y=\log _{b} x \Rightarrow y^{\prime}=1 /(x \ln b)$ (f) $y=\sin x \Rightarrow y^{\prime}=\cos x$ (g) $y=\cos x \Rightarrow y^{\prime}=-\sin x$ (h) $y=\tan x \Rightarrow y=\sec ^{2} x$ (i) $y=\csc x \Rightarrow y^{\prime}=-\operatorname{esc} x \cot x$ (j) $y=\sec x \Rightarrow y^{\prime}=\sec x \tan x$ (k) $y=\cot x \Rightarrow y^{\prime}=-\csc ^{2} x$ (l) $y=\sin ^{-1} x=y^{\prime}=1 / \sqrt{1-x^{2}}$ (m) $y=\cos ^{-1} x \Rightarrow y^{\prime}=-1 / \sqrt{1-x^{2}}$ (n) $y=\tan ^{-1} x \Rightarrow \quad y^{\prime}=1 /\left(1+x^{2}\right)$ (0) $y=\sinh x \Rightarrow y^{\prime}=\cosh x$ (p) $y=\cosh x \Rightarrow y^{\prime}=\sinh x$ (q) $y=\tanh x \Rightarrow y^{\prime}=\operatorname{sech}^{2} x$ (r) $y=\sinh ^{-1} x \Rightarrow y^{\prime}=1 / \sqrt{1+x^{2}}$ (5) $y=\cosh ^{-1} x \Rightarrow y^{\prime}=1 / \sqrt{x^{2}-1}$ (t) $y=\tanh ^{-1} x \Rightarrow y^{\prime}=1 /\left(1-x^{2}\right)$
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