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#