# Question dcd68

Oct 2, 2015

$\mathrm{dz} = 2 x \mathrm{dx} - \frac{2}{y} ^ 3 \mathrm{dy}$

#### Explanation:

z(x;y)=1/y^2+x^2-1

$\rightarrow \mathrm{dz} = \frac{\partial z}{\partial x} \mathrm{dx} + \frac{\partial z}{\partial y} \mathrm{dy}$

$\frac{\partial z}{\partial x}$ is calculated as the derivative of z(x;y)# by $x$ assuming that $y$ is constant.

$\frac{\partial z}{\partial x} = \cancel{\frac{d \left(\frac{1}{y} ^ 2\right)}{\mathrm{dx}}} + {\mathrm{dx}}^{2} / \mathrm{dx} - \cancel{\frac{d \left(1\right)}{\mathrm{dx}}} = 2 x$

Same thing for $\frac{\partial z}{\partial y}$:

$\frac{\partial z}{\partial y} = \frac{d \left(\frac{1}{y} ^ 2\right)}{\mathrm{dy}} + \cancel{{\mathrm{dx}}^{2} / \mathrm{dy}} - \cancel{\frac{d \left(1\right)}{\mathrm{dy}}} = - \frac{2}{y} ^ 3$

Therefore: $\mathrm{dz} = 2 x \mathrm{dx} - \frac{2}{y} ^ 3 \mathrm{dy}$