# Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6801

Exercise

Given the differentiable function

$$u=f(\frac{y}{x},\frac{y}{z})+\frac{xy}{z}$$

Prove the equation

$$xu'_x+yu'_y+zu'_z=\frac{xy}{z}$$

Proof

Define

$$v=\frac{y}{x}$$

$$w=\frac{y}{z}$$

We get the function

$$u=f(v,w)+\frac{xy}{z}$$

And the internal functions

$$v(x,y)=\frac{y}{x}$$

$$w(y,z)=\frac{y}{z}$$

We will use the chain rule to calculate the partial derivatives of u.

$$u'_x=f'_v\cdot v'_x+f'_w\cdot w'_x+\frac{y}{z}=$$

$$=f'_v\cdot (-\frac{y}{x^2})+f'_w\cdot 0+\frac{y}{z}=$$

$$=-\frac{y}{x^2}f'_v+\frac{y}{z}$$

$$u'_y=f'_v\cdot v'_y+f'_w\cdot w'_y+\frac{x}{z}=$$

$$=f'_v\cdot \frac{1}{x}+f'_w\cdot \frac{1}{z}+\frac{x}{z}=$$

$$=\frac{1}{x}f'_v + \frac{1}{z}f'_w+\frac{x}{z}$$

$$u'_z=f'_v\cdot v'_z+f'_w\cdot w'_z-\frac{xy}{z^2}=$$

$$=f'_v\cdot 0+f'_w\cdot (-\frac{y}{z^2})-\frac{xy}{z^2}=$$

$$=-\frac{y}{z^2}f'_w-\frac{xy}{z^2}$$

We will put the partial derivatives in the left side of the equation we need to prove.

$$xu'_x+yu'_y+zu'_z=$$

$$=x(-\frac{y}{x^2}f'_v+\frac{y}{z})+y(\frac{1}{x}f'_v + \frac{1}{z}f'_w+\frac{x}{z})+z(-\frac{y}{z^2}f'_w-\frac{xy}{z^2})=$$

$$=-\frac{y}{x}f'_v+\frac{xy}{z}+\frac{y}{x}f'_v+\frac{y}{z}f'_w+\frac{xy}{z}-\frac{y}{z}f'_w-\frac{xy}{z}=$$

$$=\frac{xy}{z}$$

We got to the right side of the equation 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!

Share with Friends