# 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: 39

$$\frac{{dy}}{{dx}} = \left[ {\left( {{x^2} + \sqrt x } \right)\left( {\ln 3} \right) + 2x + \frac{1}{{2\sqrt x }}} \right]{3^x}$$

#### Work Step by Step

\eqalign{ & y = \left( {{x^2} + \sqrt x } \right){3^x} \cr & {\text{Find }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} = \frac{{dy}}{{dx}}\left[ {\left( {{x^2} + \sqrt x } \right){3^x}} \right] \cr & {\text{By the product rule}} \cr & \frac{{dy}}{{dx}} = \left( {{x^2} + \sqrt x } \right)\frac{d}{{dx}}\left[ {{3^x}} \right] + {3^x}\frac{d}{{dx}}\left[ {{x^2} + \sqrt x } \right] \cr & {\text{differentiate}} \cr & \frac{{dy}}{{dx}} = \left( {{x^2} + \sqrt x } \right)\left( {{3^x}\ln 3} \right) + {3^x}\left( {2x + \frac{1}{{2\sqrt x }}} \right) \cr & {\text{Factor out }}{3^x} \cr & \frac{{dy}}{{dx}} = \left[ {\left( {{x^2} + \sqrt x } \right)\left( {\ln 3} \right) + 2x + \frac{1}{{2\sqrt x }}} \right]{3^x} \cr}

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