## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 3 - Derivatives - 3.4 The Product and Quotient Rules - 3.4 Exercises - Page 161: 63

#### Answer

$f'(x)=xe^{3x}(3x+2)$ $f''(x)=e^{3x}(9x^{2}+12x+2)$ $f'''(x)=9e^{3x}(3x^{2}+6x+2)$

#### Work Step by Step

$f(x)=x^{2}e^{3x}$ Start the differentiation process by using the product rule: $f'(x)=(x^{2})(e^{3x})'+(x^{2})'(e^{3x})=...$ Evaluate the derivatives indicated: $...=3x^{2}e^{3x}+2xe^{3x}=...$ Take out common factor $xe^{3x}$ and simplify: $...=xe^{3x}(3x+2)$ The expression above is the first derivative. Apply the product rule again to begin the process of finding the second derivative: $f''(x)=(xe^{3x})(3x+2)'+(3x+2)(xe^{3x})'=...$ Evaluate the derivatives indicated. Use the product rule to evaluate $(xe^{3x})'$: $...=(xe^{3x})(3)+(3x+2)[x(e^{3x})'+(x)'e^{3x}]=...$ Evaluate the remaining indicated derivatives: $...=3xe^{3x}+(3x+2)(3xe^{3x}+e^{3x})=...$ Take out common factor $e^{3x}$ from $(3xe^{3x}+e^{3x})$: $...=3xe^{3x}+e^{3x}(3x+2)(3x+1)=...$ Take out common factor $e^{3x}$ and simplify: $...=e^{3x}[3x+(3x+2)(3x+1)]=e^{3x}(9x^{2}+12x+2)$ The expression above is the second derivative. Apply the product rule one more time to begin the process of finding the third derivative: $f'''(x)=e^{3x}(9x^{2}+12x+2)'+(9x^{2}+12x+2)(e^{3x})'=...$ Evaluate the indicated derivatives: $...=e^{3x}(18x+12)+3e^{3x}(9x^{2}+12x+2)=...$ Take out common factor $e^{3x}$ and simplify: $...=e^{3x}[3(9x^{2}+12x+2)+18x+12]=...$ $...=e^{3x}(27x^{2}+54x+18)=9e^{3x}(3x^{2}+6x+2)$ The expression above is the third derivative.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.