## Calculus 8th Edition

Published by Cengage

# Chapter 14 - Partial Derivatives - Review - Exercises: 13

#### Answer

$f_{x}=32xy(5y^{3}+2x^{2}y)^{7}$ and $f_{y}=8(5y^{3}+2x^{2}y)^{7} (15y^{2}+2x^{2})$

#### Work Step by Step

Given: $f(x,y)=(5y^{3}+2x^{2}y)^{8}$ Need to find first partial derivatives $f_{x}$ and $f_{y}$. Differentiate the function with respect to $x$ keeping $y$ constant. $f_{x}=8(5y^{3}+2x^{2}y)^{8-1}\times4xy$ $f_{x}=32xy8(5y^{3}+2x^{2}y)^{7}$ Differentiate the function with respect to $y$ keeping $x$ constant. $f_{y}=8(5y^{3}+2x^{2}y)^{8-1}\times (15y^{2}+2x^{2})$ Hence, $f_{x}=32xy(5y^{3}+2x^{2}y)^{7}$ and $f_{y}=8(5y^{3}+2x^{2}y)^{7} (15y^{2}+2x^{2})$

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