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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