Exercise
Given the differentiable function
u(x,y,z)=f(x^2z-yz)
Prove the equation
xu'_x+2yu'_y-2zu'_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
t=x^2z-yz
We get the function
u(x,y,z)=f(t)
Or simply
u(x,y,z)=f
Now we will use the chain rule to calculate the partial derivatives of z.
u'_x=f'_t\cdot t'_x
u'_y=f'_t\cdot t'_y
u'_z=f'_t\cdot t'_z
We will calculate the partial derivatives of t.
t'_x=2xz
t'_y=-z
t'_z=x^2-y
We put the results in the partial derivatives of u and get
u'_x=f'_t\cdot t'_x=f'_t\cdot 2xz
u'_y=f'_t\cdot t'_y=f'_t\cdot (-z)
u'_z=f'_t\cdot t'_z=f'_t\cdot (x^2-y)
We will put the partial derivatives in the left side of the equation we need to prove.
xu'_x+2yu'_y-2zu'_z=
=xf'_t\cdot 2xz+2yf'_t\cdot (-z)-2zf'_t\cdot (x^2-y)=
=2zx^2f'_t-2yzf'_t-2zx^2f'_t +2yzf'_t=0
Hence, we get
xu'_x+2yu'_y-2zu'_z=0
As required.
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