Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.3 Derivatives Of Inverse Functions; Derivatives And Integrals Involving Exponential Functions - Exercises Set 6.3 - Page 432: 41

Answer

$$\frac{{dy}}{{dx}} = {4^{3\sin x - {e^x}}}\ln 4\left( {3\cos x - {e^x}} \right)$$

Work Step by Step

$$\eqalign{ & y = {4^{3\sin x - {e^x}}} \cr & {\text{Find }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{4^{3\sin x - {e^x}}}} \right] \cr & {\text{use the formula }}\frac{d}{{dx}}\left[ {{b^u}} \right] = {b^u}\ln b \cdot \frac{{du}}{{dx}}\,\,\,\left( {See\,\,page\,\,\,429} \right) \cr & {\text{consider }}b = 4{\text{ and }}u = 3\sin x - {e^x}.\,{\text{Thus}}{\text{,}} \cr & \frac{{dy}}{{dx}} = {4^{3\sin x - {e^x}}}\ln 4 \cdot \frac{d}{{dx}}\left[ {3\sin x - {e^x}} \right] \cr & {\text{differentiate}} \cr & \frac{{dy}}{{dx}} = {4^{3\sin x - {e^x}}}\ln 4\left( {3\cos x - {e^x}} \right) \cr} $$
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