Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.4 Derivatives of Logarithmic Functions - 6.4 Exercises: 44

Answer

$y'=\frac{e^{-x}cos^{2}x}{x^{2}+x+1}[-2tanx-1-\frac{2x+1}{{x^{2}+x+1}}]$

Work Step by Step

Use logarithmic differentiation to find the derivative of the function $y=\frac{e^{-x}cos^{2}x}{x^{2}+x+1}$ Taking logarithmic on both sides . $lny=ln[\frac{e^{-x}cos^{2}x}{x^{2}+x+1}]$ Use logarithmic properties $ln(xy)=lnx+lny$ , $ln(\frac{x} {y})=lnx-lny$ and $ln(x^{y})=ylnx$. $lny=ln[{e^{-x}cos^{2}x}]-ln[{x^{2}+x+1}]$ Differentiate with respect to $x$. $\frac{y'}{y}=\frac{1}{{e^{-x}cos^{2}x}}\frac{d}{dx}({e^{-x}cos^{2}x})-\frac{1}{{x^{2}+x+1}}\frac{d}{dx}({x^{2}+x+1})$ $=\frac{(e^{-x}-2cosxsinx)-(e^{-x}cos^{2}x)}{{e^{-x}cos^{2}x}}-\frac{2x+1}{{x^{2}+x+1}}$ $=\frac{-sin2x-cos^{2}x}{{cos^{2}x}}-\frac{2x+1}{{x^{2}+x+1}}$ $=-2tanx-1-\frac{2x+1}{{x^{2}+x+1}}$ Thus, $y'=y[-2tanx-1-\frac{2x+1}{{x^{2}+x+1}}]$ Hence, $y'=\frac{e^{-x}cos^{2}x}{x^{2}+x+1}[-2tanx-1-\frac{2x+1}{{x^{2}+x+1}}]$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.