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: 38

Answer

$$p'\left( x \right) = - 2{x^{ - \ln x - 1}}\ln x$$

Work Step by Step

$$\eqalign{ & p\left( x \right) = {x^{ - \ln x}} \cr & {\text{we use the inverse relationship }}{e^{\ln x}} = x \cr & p\left( x \right) = {e^{\ln {x^{ - \ln x}}}} \cr & p\left( x \right) = {e^{ - \ln x\ln x}} \cr & p\left( x \right) = {e^{ - {{\left( {\ln x} \right)}^2}}} \cr & {\text{evaluate the derivative}} \cr & p'\left( x \right) = \frac{d}{{dx}}\left( {{e^{ - {{\left( {\ln x} \right)}^2}}}} \right) \cr & {\text{by }}\frac{d}{{dx}}\left( {{e^{u\left( x \right)}}} \right) = {e^{u\left( x \right)}}u'\left( x \right) \cr & p'\left( x \right) = {e^{ - {{\left( {\ln x} \right)}^2}}}\frac{d}{{dx}}\left( { - {{\left( {\ln x} \right)}^2}} \right) \cr & {\text{chain rule}} \cr & p'\left( x \right) = - {e^{ - {{\left( {\ln x} \right)}^2}}}\left( {2\ln x} \right)\left( {\frac{1}{x}} \right) \cr & {\text{simplify}} \cr & p'\left( x \right) = \frac{{ - 2{x^{ - \ln x}}\ln x}}{x} \cr & p'\left( x \right) = - 2{x^{ - \ln x - 1}}\ln x \cr} $$
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