# Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6458

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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