Find $\frac{\partial z}{\partial x}$ from these equation:
$xyz-2xz+3yz-4xy=0$
How $\frac{\partial z}{\partial x}$ can be done using manually method? Thanks... I stuck at variable whose contains 'z'...
$\endgroup$1 Answer
$\begingroup$Denote $\dfrac{\partial z}{\partial x} = z'$, we have: $xyz - 2xz + 3yz - 4xy = 0 \Rightarrow z(xy-2x+3y) = 4xy$. Using the product rule and differentiating both sides with respect to $x$:
$z'(xy-2x+3y) + z(y-2) = 4y \Rightarrow \dfrac{4xyz'}{z} = 4y - yz + 2z \Rightarrow z' = \dfrac{z(4y-yz+2z)}{4xy} = \dfrac{z(4y-yz+2z)}{z(xy-2x+3y)} = \dfrac{-yz+2z+4y}{3y-2x+yx}$
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