## Calculus 8th Edition

$\frac{\partial z}{\partial y}=\frac{xz}{e^z -xy}$ $\frac{\partial z}{\partial x}=\frac{yz}{e^z -xy}$
The equation $e^z=xyz$ is given. To find its partial derivatives, one must differentiate both sides of the equation with respect to either $x$ or $y$ while maintaining that $z$ is an implicit function of either $x$ or $y$. To find the partial derivative $f_x$: $\frac{\partial}{\partial x} {e^z} = \frac{\partial}{\partial x} {xyz}$ ${e^z} \frac{\partial z}{\partial x}= \frac{\partial z}{\partial x}{xy}+yz$ Seperating the partial derivative: $\frac{\partial z}{\partial x}=\frac{yz}{e^z -xy}$ To find the partial derivative $f_x$: $\frac{\partial}{\partial y} {e^z} = \frac{\partial}{\partial y} {xyz}$ ${e^z} \frac{\partial z}{\partial x}= \frac{\partial z}{\partial y}{xy}+yxz$ Seperating the partial derivative: $\frac{\partial z}{\partial x}=\frac{yz}{e^z -xy}$