Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 13 - Functions of Several Variables - 13.3 Exercises - Page 896: 42

Answer

$${f_x}\left( {x,y} \right) = - \frac{1}{{{{\left( {x + y} \right)}^2}}}{\text{ and }}{f_y}\left( {x,y} \right) = - \frac{1}{{{{\left( {x + y} \right)}^2}}}$$

Work Step by Step

$$\eqalign{ & f\left( {x,y} \right) = \frac{1}{{x + y}} \cr & {\text{Differentiate by using the limit definition}} \cr & {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h,y} \right) - f\left( {x,y} \right)}}{h} \cr & {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{x + h + y}} - \frac{1}{{x + y}}}}{h} \cr & {\text{Rationalizing}} \cr & {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{x + y - x - h - y}}{{\left( {x + h + y} \right)\left( {x + y} \right)}}}}{h} \cr & {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{ - h}}{{\left( {x + h + y} \right)\left( {x + y} \right)}}}}{h} \cr & {f_x}\left( {x,y} \right) = - \mathop {\lim }\limits_{h \to 0} \frac{1}{{\left( {x + h + y} \right)\left( {x + y} \right)}} \cr & {\text{Evaluate the limit}} \cr & {f_x}\left( {x,y} \right) = - \frac{1}{{\left( {x + 0 + y} \right)\left( {x + y} \right)}} \cr & {f_x}\left( {x,y} \right) = - \frac{1}{{{{\left( {x + y} \right)}^2}}} \cr & \cr & {\text{Differentiate by using the limit definition}} \cr & {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{x + y + h}} - \frac{1}{{x + y}}}}{h} \cr & {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{x + y - x - y - h}}{{\left( {x + y + h} \right)\left( {x + y} \right)}}}}{h} \cr & {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{ - h}}{{\left( {x + y + h} \right)\left( {x + y} \right)}}}}{h} \cr & {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{ - h}}{{h\left( {x + y + h} \right)\left( {x + y} \right)}} \cr & {f_y}\left( {x,y} \right) = - \mathop {\lim }\limits_{h \to 0} \frac{1}{{\left( {x + y + h} \right)\left( {x + y} \right)}} \cr & {\text{Evaluate the limit}} \cr & {f_y}\left( {x,y} \right) = - \frac{1}{{\left( {x + y} \right)\left( {x + y} \right)}} \cr & {f_y}\left( {x,y} \right) = - \frac{1}{{{{\left( {x + y} \right)}^2}}} \cr} $$
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