Question #dcd68

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Answer 1 · 2015-10-02T21:15:43

$d z = 2 x d x - \frac{2}{y^{3}} d y$ $dz=2xdx-2/y^3dy$

$z (x; y) = \frac{1}{y^{2}} + x^{2} - 1$ $z(x;y)=1/y^2+x^2-1$

$\to d z = \frac{\partial z}{\partial x} d x + \frac{\partial z}{\partial y} d y$ $rarr dz=(delz)/(delx)dx+(delz)/(dely)dy$

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

$(delz)/(delx)=cancel((d(1/y^2))/dx)+dx^2/dx-cancel((d(1))/dx)=2x$

Same thing for $(delz)/(dely)$ :

$(delz)/(dely)=(d(1/y^2))/dy+cancel(dx^2/dy)-cancel((d(1))/dy)=-2/y^3$

Therefore: $dz=2xdx-2/y^3dy$