Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

Chapter 6 - Applications of Integration - 6.8 Logarithmic and Exponential - 6.8 Exercises: 7

Answer

$$3$$

Work Step by Step

\eqalign{ & {\left. {\frac{d}{{dx}}\left( {x\ln {x^3}} \right)} \right|_{x = 1}} \cr & {\text{use product rule}} \cr & = x\frac{d}{{dx}}\left( {\ln {x^3}} \right) + \ln {x^3}\frac{d}{{dx}}\left( x \right) \cr & = x\left( {\frac{{3{x^2}}}{{{x^3}}}} \right) + \ln {x^3}\left( 1 \right) \cr & = 3 + \ln {x^3} \cr & {\text{evaluate the derivative at }}x = 1 \cr & = 3 + \ln {\left( 1 \right)^3} \cr & {\text{Simplify}} \cr & = 3 \cr}

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