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: 36

Answer

$$h'\left( t \right) = \sqrt t {\left( {\sin t} \right)^{\sqrt t }}\cos t + \frac{{{{\left( {\sin t} \right)}^{\sqrt t + 1}}}}{{2\sqrt t }}$$

Work Step by Step

$$\eqalign{ & h\left( t \right) = {\left( {\sin t} \right)^{\sqrt t }} \cr & {\text{we use the inverse relationship }}{e^{\ln x}} = x \cr & h\left( t \right) = {e^{{{\left( {\sin t} \right)}^{\sqrt t }}}} \cr & h\left( t \right) = {e^{\sqrt t \sin t}} \cr & {\text{evaluate the derivative}} \cr & h'\left( t \right) = \frac{d}{{dt}}\left( {{e^{\sqrt t \sin t}}} \right) \cr & {\text{by }}\frac{d}{{dx}}\left( {{e^{u\left( x \right)}}} \right) = {e^{u\left( x \right)}}u'\left( x \right) \cr & h'\left( t \right) = {e^{\sqrt t \sin t}}\frac{d}{{dt}}\left( {\sqrt t \sin t} \right) \cr & {\text{product rule}} \cr & h'\left( t \right) = {e^{\sqrt t \sin t}}\left( {\sqrt t \cos t + \frac{{\sin t}}{{2\sqrt t }}} \right) \cr & {\text{simplify}} \cr & h'\left( t \right) = {\left( {\sin t} \right)^{\sqrt t }}\left( {\sqrt t \cos t + \frac{{\sin t}}{{2\sqrt t }}} \right) \cr & h'\left( t \right) = \sqrt t {\left( {\sin t} \right)^{\sqrt t }}\cos t + \frac{{{{\left( {\sin t} \right)}^{\sqrt t + 1}}}}{{2\sqrt t }} \cr} $$
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