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Q12E

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Essential Calculus: Early Transcendentals
Found in: Page 656
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

Determine the values of the derivatives \({g_r}(1,2)\)and \({g_r}(1,2).\)The functions are \(g(r,s) = f\left( {2r - s,{s^2} - 4r} \right),x = x(r,s)\) and \(y = y(r,s).\)

The values of derivatives are \({g_r}(1,2) = - 24\)and\({g_s}(1,2) = 28\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: The chain rule.

With the chain rule, the partial derivative of function \({\rm{f(x, y)}}\) with respect to \({\rm{t}}\) becomes \(\frac{{\partial f}}{{\partial t}} = \frac{{\partial f}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial t}} + \frac{{\partial f}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial t}}.\)

Step 2: Use the chain rule for calculation.

Differentiate \(g\) partially with respect to \(r\) as follows.

\(\begin{aligned}{l}\frac{{\partial g}}{{\partial r}} = \frac{\partial }{{\partial r}}f(x(r,s),y(r,s))\\ = \frac{{\partial f}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial r}} + \frac{{\partial f}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial r}}\\ = {f_x}(x(r,s),y(r,s)) \cdot {x_r}(r,s) + {f_y}(x(r,s),y(r,s)) \cdot {y_r}(r,s)\\ = {f_x}(x(r,s),y(r,s)) \cdot 2 + {f_y}(x(r,s),y(r,s)) \cdot ( - 4)\end{aligned}\)

At \(r = 1\)and\(s = 2\), the partial derivatives becomes,

\(\begin{aligned}{l}{\left. {\frac{{\partial g}}{{\partial r}}} \right|_{r = 1,s = 2}} = 2{f_x}\left( {2(1) - (2),{{(2)}^2} - 4(1)} \right) - 4{f_y}\left( {2(1) - (2),{{(2)}^2} - 4(1)} \right)\\ = 2{f_x}(0,0) - 4{f_y}(0,0)\\ = 2(4) - 4(8)\\ = - 24\end{aligned}\)

That is, \({g_r}(1,2) = - 24\).

Differentiate \(g\) partially with respect to \(s\) as follows.

\(\begin{aligned}{l}\frac{{\partial g}}{{\partial s}} = \frac{\partial }{{\partial s}}f(x(r,s),y(r,s))\\ = \frac{{\partial f}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial s}} + \frac{{\partial f}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial s}}\\ = {f_x}(x(r,s),y(r,s)) \cdot {x_s}(r,s) + {f_y}(x(r,s),y(r,s)) \cdot {y_s}(r,s)\\ = {f_x}(x(r,s),y(r,s)) \cdot ( - 1) + {f_y}(x(r,s),y(r,s)) \cdot (2s)\end{aligned}\)

At \(r = 1\) and\(s = 2\), the partial derivatives becomes,

\(\begin{aligned}{l}{\left. {\frac{{\partial g}}{{\partial s}}} \right|_{r = 1,s = 2}} = - {f_x}\left( {2(1) - (2),{{(2)}^2} - 4(1)} \right) + 2(2){f_y}\left( {2(1) - (2),{{(2)}^2} - 4(1)} \right)\\ = - {f_x}(0,0) + 4{f_y}(0,0)\\ = - (4) + 4(8)\\ = 28\end{aligned}\)

That is, \({g_s}(1,2) = 28\).

Therefore, the values of derivatives are \({g_r}(1,2) = - 24\)and \({g_s}(1,2) = 28\).

Most popular questions for Math Textbooks

Graph and discuss the continuity of the function

\(f(x,y) = \left\{ \begin{aligned}{l}\frac{{sinxy}}{{xy}}, if xy \ne 0\\1, if xy = 0\end{aligned} \right.\)

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