# Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6504

Exercise

Given the differentiable function

$$u=f(x-y,y-z,z-x)$$

Prove the equation

$$u'_x+u'_y+u'_z=0$$

Proof

Define

$$v=x-y$$

$$w=y-z$$

$$k=z-x$$

We get the function

$$u=f(v,w,k)$$

And the internal functions

$$v(x,y)=x-y$$

$$w(y,z)=y-z$$

$$k(z,x)=z-x$$

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

$$u'_x=f'_v\cdot v'_x+f'_w\cdot w'_x+f'_k\cdot k'_x=$$

$$=f'_v\cdot 1+f'_w\cdot 0+f'_k\cdot (-1)=$$

$$=f'_v-f'_k$$

$$u'_y=f'_v\cdot v'_y+f'_w\cdot w'_y+f'_k\cdot k'_y=$$

$$=f'_v\cdot (-1)+f'_w\cdot 1+f'_k\cdot 0=$$

$$=f'_w-f'_v$$

$$u'_z=f'_v\cdot v'_z+f'_w\cdot w'_z+f'_k\cdot k'_z=$$

$$=f'_v\cdot 0+f'_w\cdot (-1)+f'_k\cdot 1=$$

$$=f'_k-f'_w$$

We got the equations:

$$u'_x=f'_v-f'_k$$

$$u'_y=f'_w-f'_v$$

$$u'_z=f'_k-f'_w$$

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

$$u'_x+u'_y+u'_z=$$

$$=f'_v-f'_k+f'_w-f'_v+f'_k-f'_w=$$

$$=0$$

We got zero as required.

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