Question #dcd68

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Answer 1 · 2015-10-02T21:15:43

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

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

$rarr dz=(delz)/(delx)dx+(delz)/(dely)dy$

$(delz)/(delx)$ is calculated as the derivative of $z(x;y)$ by $x$ assuming that $y$ 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: $dz=2xdx-2/y^3dy$