Exercise
Given the differentiable function
U(x,y,z)=x+\frac{x-y}{y-z}
Prove the equation
U'_x+U'_y+U'_z=1
Proof
We will use the chain rule to calculate the partial derivatives of U.
U'_x=1+\frac{1}{y-z}
U'_y=\frac{-(y-z)-(x-y)}{{(y-z)}^2}=
=\frac{z-x}{{(y-z)}^2}
U'_z=\frac{-(x-y)\cdot (-1)}{{(y-z)}^2}=
=\frac{x-y}{{(y-z)}^2}
We will put the partial derivatives in the left side of the equation we need to prove.
U'_x+U'_y+U'_z=
=1+\frac{1}{y-z}+\frac{z-x}{{(y-z)}^2}+\frac{x-y}{{(y-z)}^2}=
=1+\frac{1}{y-z}+\frac{z-x+x-y}{{(y-z)}^2}=
=1+\frac{1}{y-z}+\frac{z-y}{{(y-z)}^2}=
=1+\frac{1}{y-z}-\frac{y-z}{{(y-z)}^2}=
=1+\frac{1}{y-z}-\frac{1}{y-z}=
=1
We got one as required.
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