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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