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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