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.

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions!