Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

\({e^z} = xyz\)

Short Answer

Expert verified

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

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

Step by step solution

01

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\({z^{\prime \prime }}\).

02

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free