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Found in: Page 656

### Essential Calculus: Early Transcendentals

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

# 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$$\frac{{xz}}{{{e^z} - xy}}$$.

See the step by step solution

## 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.