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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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