## Calculus 8th Edition

$\frac{\partial z}{\partial x}=\frac{-x}{z-1}$ $\frac{\partial z}{\partial y}=\frac{y}{z-1}$
Use implicit differentiation to find the partial derivatives of the following function: $x^2-y¨2+z^2-2z=4$ We first differentiate both sides of the equation with respect to $x$. All constants become zero: $2x-0+2z\frac{\partial z}{\partial x}-2\frac{\partial z}{\partial x}=0$ $\frac{\partial z}{\partial x}=\frac{-2x}{2z-2}=\frac{-x}{z-1}$ We now differentiate both sides of the equation with respect to $y$, just as done above: $0-2y+2z\frac{\partial z}{\partial y}-2\frac{\partial z}{\partial y}=0$ $\frac{\partial z}{\partial y}=\frac{2y}{2z-2}=\frac{y}{z-1}$