Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

Find the value of \(\frac{{\partial z}}{{\partial x}}\) and \(\frac{{\partial z}}{{\partial y}}\) using equation 7.
\({e^z} = xyz\)

The value of \(\frac{{\partial z}}{{\partial x}}\) is\(\frac{{yz}}{{{e^z} - xy}}\).

The value of \(\frac{{\partial z}}{{\partial y}}\) is\(\frac{{xz}}{{{e^z} - xy}}\).

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Formula used

Equation 7: " \(\frac{{\partial z}}{{\partial x}} = - \frac{{\frac{{\partial F}}{{dx}}}}{{\frac{{\partial F}}{{dz}}}} = - \frac{{{F_x}}}{{{F_z}}}\) and \(\frac{{\partial z}}{{\partial y}} = - \frac{{\frac{{\partial F}}{{\partial y}}}}{{\frac{{\partial F}}{{\partial z}}}} = - \frac{{{F_y}}}{{{F_z}}}\), where \(F\) is the function of x, y and \({z^{\prime \prime }}\).

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

As given function is,\({e^z} = xyz\)

Let the function be, \(F(x,y,z) = {e^z} - xyz \ldots \ldots \ldots (1)\)

The equations of \(\frac{{\partial z}}{{\partial x}}\) and \(\frac{{\partial z}}{{\partial y}}\) using the equation 7 is as follows,

\(\begin{aligned}{l}\frac{{\partial z}}{{\partial x}} = - \frac{{{F_x}}}{{{F_z}}} \ldots \ldots \ldots (2)\\\frac{{\partial z}}{{\partial y}} = - \frac{{{F_y}}}{{{F_z}}} \ldots \ldots \ldots (3)\end{aligned}\)

Take partial derivative with respect to \(x\) in the equation (1),

\(\begin{aligned}{l}\frac{{\partial F}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {{e^z} - xyz} \right)\\{F_x} = 0 - (1)yz\\{F_x} = - yz\end{aligned}\)

Thus, the partial derivate of\(x\)is\({F_x} = - yz\).

Take the partial derivative with respect to \(y\) in the equation (1),

\(\begin{aligned}{l}\frac{{\partial F}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {{e^z} - xyz} \right)\\{F_y} = 0 - x(1)z\\{F_y} = - xz\end{aligned}\)

Thus, the partial derivate of\(y\)is\({F_y} = - xz\).

Take the partial derivative with respect to \(z\) in the equation (1),

\(\begin{aligned}{l}\frac{{\partial F}}{{\partial z}} = \frac{\partial }{{\partial z}}\left( {{e^z} - xyz} \right)\\{F_z} = {e^z} - xy \ldots \ldots \ldots (1)\\{F_z} = {e^z} - xy\end{aligned}\)

Substitute the respective values in the equation (2),

\(\begin{aligned}{l}\frac{{\partial z}}{{\partial x}} = - \frac{{{F_x}}}{{{F_z}}}\\\frac{{\partial z}}{{\partial x}} = - \frac{{ - yz}}{{{e^z} - xy}}\\\frac{{\partial z}}{{\partial x}} = \frac{{yz}}{{{e^z} - xy}}\end{aligned}\)

Therefore, the value of \(\frac{{\partial z}}{{\partial x}}\) is\(\frac{{yz}}{{{e^z} - xy}}\).

Then, substitute the respective values in the equation (3),

\(\begin{aligned}{l}\frac{{\partial z}}{{\partial y}} = - \frac{{{F_y}}}{{{F_z}}}\\\frac{{\partial z}}{{\partial y}} = - \frac{{ - xz}}{{{e^z} - xy}}\\\frac{{\partial z}}{{\partial y}} = \frac{{xz}}{{{e^{z - xy}}}}\end{aligned}\)

Therefore, the value of \(\frac{{\partial z}}{{\partial y}}\) is\(\frac{{xz}}{{{e^z} - xy}}\).

Hence, the values of \(\frac{{\partial z}}{{\partial x}}\)and \(\frac{{\partial z}}{{\partial y}}\)are \(\frac{{yz}}{{{e^z} - xy}}\)and \(\frac{{xz}}{{{e^z} - xy}}\), respectively.

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Essential Calculus: Early Transcendentals

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Short Answer

Find the value of \(\frac{{\partial z}}{{\partial x}}\) and \(\frac{{\partial z}}{{\partial y}}\) using equation 7.
\({e^z} = xyz\)

Step by Step Solution

Step 1: Formula used

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

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Find the value of \(\frac{{\partial z}}{{\partial x}}\) and \(\frac{{\partial z}}{{\partial y}}\) using equation 7.\({e^z} = xyz\)

Step by Step Solution

Step 1: Formula used

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

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Find the value of \(\frac{{\partial z}}{{\partial x}}\) and \(\frac{{\partial z}}{{\partial y}}\) using equation 7.
\({e^z} = xyz\)