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: 41

Answer

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

Work Step by Step

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