Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 6 - Applications of Integration - 6.8 Logarithmic and Exponential - 6.8 Exercises: 56

Answer

$${x^{{x^{10}} + 9}}\left( {1 + 10\ln x} \right)$$

Work Step by Step

$$\eqalign{ & \frac{d}{{dx}}\left( {{x^{\left( {{x^{10}}} \right)}}} \right) \cr & {\text{we use the inverse relationship }}{e^{\ln x}} = x \cr & = {e^{\ln {x^{\left( {{x^{10}}} \right)}}}} \cr & = {e^{{x^{10}}\ln x}} \cr & {\text{evaluate the derivative}} \cr & = \frac{d}{{dx}}\left( {{e^{{x^{10}}\ln x}}} \right) \cr & {\text{by }}\frac{d}{{dx}}\left( {{e^{u\left( x \right)}}} \right) = {e^{u\left( x \right)}}u'\left( x \right) \cr & = {e^{{x^{10}}\ln x}}\frac{d}{{dx}}\left( {{x^{10}}\ln x} \right) \cr & {\text{product rule}} \cr & = {e^{{x^{10}}\ln x}}\left( {\frac{{{x^{10}}}}{x} + 10{x^9}\ln x} \right) \cr & {\text{simplify}} \cr & = {x^{\left( {{x^{10}}} \right)}}\left( {{x^9} + 10{x^9}\ln x} \right) \cr & = {x^{{x^{10}} + 9}}\left( {1 + 10\ln x} \right) \cr} $$
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