# Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 3384

Exercise

Given the differentiable function

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

Prove the equation

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

Proof

When we have a function and in its parentheses there is a complex expression instead of a simple variable, we will define a new variable like this

$$v=x-y$$

$$w=y-z$$

$$t=z-x$$

We get the function

$$u(x,y,z)=f(v,w,t)$$

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

$$u'_x=f'_v\cdot v'_x+f'_w\cdot w'_x+f'_t\cdot t'_x$$

$$u'_y=f'_v\cdot v'_y+f'_w\cdot w'_y+f'_t\cdot t'_y$$

$$u'_z=f'_v\cdot v'_z+f'_w\cdot w'_z+f'_t\cdot t'_z$$

We calculate  the partial derivatives of v,w and t.

$$v'_x=1$$

$$v'_y=-1$$

$$v'_z=0$$

$$w'_x=0$$

$$w'_y=1$$

$$w'_z=-1$$

$$t'_x=-1$$

$$t'_y=0$$

$$t'_z=1$$

We put the results in the partial derivatives of u and get

$$u'_x=f'_v\cdot v'_x+f'_w\cdot w'_x+f'_t\cdot t'_x=$$

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

$$=f'_v-f'_t$$

$$u'_y=f'_v\cdot -1+f'_w\cdot w'_y+f'_t\cdot t'_y=$$

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

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

$$u'_z=f'_v\cdot v'_z+f'_w\cdot w'_z+f'_t\cdot t'_z=$$

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

$$=-f'_w+f'_t$$

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

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

$$=(f'_v-f'_t)+(-f'_v+f'_w)+(-f'_w+f'_t)=$$

$$=f'_v-f'_t-f'_v+f'_w-f'_w+f'_t=0$$

Hence, we got

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

As required.

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