## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 4 - Calculating the Derivative - Chapter Review - Review Exercises - Page 244: 46

#### Answer

$$\frac{{dy}}{{dx}} = \frac{{5\left( {\ln 2} \right)\left( {{2^{\sqrt x }}} \right)}}{{\sqrt x }}$$

#### Work Step by Step

\eqalign{ & y = 10 \cdot {2^{\sqrt x }} \cr & {\text{differentiate both sides}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {10 \cdot {2^{\sqrt x }}} \right] \cr & \frac{{dy}}{{dx}} = 10\frac{d}{{dx}}\left[ {{2^{\sqrt x }}} \right] \cr & {\text{use the derivative of }}{a^{g\left( x \right)}},\,\,\,\,\,\frac{d}{{dx}}\left[ {{a^{g\left( x \right)}}} \right] = \left( {\ln a} \right){a^{g\left( x \right)}}g'\left( x \right) \cr & \frac{{dy}}{{dx}} = 10\left( {{2^{\sqrt x }}} \right)\left( {\ln 2} \right)\frac{d}{{dx}}\left[ {\sqrt x } \right] \cr & \frac{{dy}}{{dx}} = 10\left( {{2^{\sqrt x }}} \right)\left( {\ln 2} \right)\left( {\frac{1}{{2\sqrt x }}} \right) \cr & {\text{simplifying}} \cr & \frac{{dy}}{{dx}} = \frac{{5\left( {\ln 2} \right)\left( {{2^{\sqrt x }}} \right)}}{{\sqrt x }} \cr}

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