University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 175: 47



Work Step by Step

$$y=t(t+1)(t+2)$$ Logarithmic differentiation helps to simplify the differentiation of otherwise complex functions by using the algebraic properties of logarithm. In detail, 1) Take the natural logarithm of both sides and simplify: $$\ln y=\ln\Big(t(t+1)(t+2)\Big)$$ Recall that $\ln(A\times B)=\ln A+\ln B$: $$\ln y=\ln t+\ln(t+1)+\ln(t+2)$$ 2) Then we take the derivative of both sides with respect to $t$ and remember that $(\ln t)'=1/t$: $$\frac{1}{y}\times y'=\frac{1}{t}+\frac{1}{t+1}(t+1)'+\frac{1}{t+2}(t+2)'$$ $$\frac{y'}{y}=\frac{1}{t}+\frac{1}{t+1}+\frac{1}{t+2}$$ $$\frac{y'}{y}=\frac{(t+1)(t+2)+t(t+2)+t(t+1)}{t(t+1)(t+2)}$$ $$\frac{y'}{y}=\frac{(t^2+3t+2)+(t^2+2t)+(t^2+t)}{t(t+1)(t+2)}$$ $$\frac{y'}{y}=\frac{3t^2+6t+2}{t(t+1)(t+2)}$$ 3) Solve for $y'$: $$y'=\frac{3t^2+6t+2}{t(t+1)(t+2)}\times y$$ 4) Finally, substitute for $y=t(t+1)(t+2)$ $$y'=\frac{3t^2+6t+2}{t(t+1)(t+2)}\times t(t+1)(t+2)=3t^2+6t+2$$
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.