# Chapter 7: Transcendental Functions - Section 7.3 - Exponential Functions - Exercises 7.3 - Page 390: 11

$$\frac{{dy}}{{dx}} = {x^2}{e^x}$$

#### Work Step by Step

\eqalign{ & y = \left( {{x^2} - 2x + 2} \right){e^x} \cr & {\text{Find the derivative of }}y{\text{ with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\left( {{x^2} - 2x + 2} \right){e^x}} \right] \cr & {\text{use the product rule }} \cr & \frac{{dy}}{{dx}} = \left( {{x^2} - 2x + 2} \right)\frac{d}{{dx}}\left[ {{e^x}} \right] + {e^x}\frac{d}{{dx}}\left[ {{x^2} - 2x + 2} \right] \cr & {\text{solve the derivatives and simplify}} \cr & \frac{{dy}}{{dx}} = \left( {{x^2} - 2x + 2} \right)\left( {{e^x}} \right) + {e^x}\left( {2x - 2} \right) \cr & \frac{{dy}}{{dx}} = \left( {{x^2} - 2x + 2} \right){e^x} + {e^x}\left( {2x - 2} \right) \cr & {\text{factor}} \cr & \frac{{dy}}{{dx}} = \left( {{x^2} - 2x + 2 + 2x - 2} \right){e^x} \cr & \frac{{dy}}{{dx}} = {x^2}{e^x} \cr}

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