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Q39E

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Essential Calculus: Early Transcendentals
Found in: Page 657
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

Show the equation \(\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}} = 0\) holds true.

The equation \(\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}} = 0\) holds true.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Formula used

Equation 7: " \(\frac{{\partial z}}{{\partial x}} = - \frac{{\frac{{\partial F}}{{dx}}}}{{\frac{{\partial F}}{{dz}}}}\) and \(\frac{{\partial z}}{{\partial y}} = - \frac{{\frac{{\partial F}}{{\partial y}}}}{{\frac{{\partial F}}{{\partial x}}}}\)

The function \(F\) becomes\(F(x) = z - f(x - y)\).

Step 2: Determine the partial derivatives of \(\frac{{\partial z}}{{\partial x}}\)and \(\frac{{\partial z}}{{\partial y}}\) 

As given that the function is \(z = f(x - y)\)

That is, \(z - f(x - y) = 0\)

Obtain the value of \(\frac{{\partial z}}{{\partial x}}\) as follows

\(\begin{aligned}{l}\frac{{\partial z}}{{\partial x}} = - \frac{{\frac{\partial }{{\partial x}}(z - f(x - y))}}{{\frac{\partial }{\partial }(z - f(x - y))}}\\\frac{{\partial z}}{{\partial x}} = \frac{{ - {f^\prime }(x - y)}}{1}\\\frac{{\partial z}}{{\partial x}} = - {f^\prime }(x - y)\end{aligned}\)

Obtain the value of \(\frac{{\partial z}}{{\partial y}}\) as follows,

\(\begin{aligned}{l}\frac{{\partial z}}{{\partial y}} = - \frac{{\frac{\partial }{{\partial y}}(z - f(x - y))}}{{\frac{\partial }{\partial }(z - f(x - y))}}\\\frac{{\partial z}}{{\partial y}} = \frac{{ - {f^\prime }(x - y)( - 1)}}{1}\\\frac{{\partial z}}{{\partial y}} = {f^\prime }(x - y)\end{aligned}\)

Thus, the sum of the partial derivatives becomes\(\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}} = - {f^\prime }(x - y) + {f^\prime }(x - y) = 0\). Hence, the equation \(\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}} = 0\) holds true.

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