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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