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: 52

Answer

$${x^{\tan x}}\left( {\frac{{\tan x}}{x} + {{\sec }^2}x\ln x} \right)$$

Work Step by Step

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