Exercise
Given the differentiable function
z=\frac{y^2}{3x}+f(xy)
Prove the equation
x^2z'_x-xyz'_y+y^2=0
Proof
Define
u=xy
We get the function
z=\frac{y^2}{3x}+f(u)
And the internal function
u(x,y)=xy
We will use the chain rule to calculate the partial derivatives of z.
z'_x=\frac{y^2}{3}\cdot (-\frac{1}{x^2})+f'_u\cdot u'_x=
=-\frac{y^2}{3x^2}+f'_u\cdot y
z'_y=\frac{1}{3x}\cdot 2y+f'_u\cdot u'_y=
=\frac{2y}{3x}+f'_u\cdot x
We will put the partial derivatives in the left side of the equation we need to prove.
x^2z'_x-xyz'_y+y^2=
=x^2(-\frac{y^2}{3x^2}+yf'_u)-xy(\frac{2y}{3x}+xf'_u)+y^2=
=-\frac{y^2}{3}+x^2yf'_u-\frac{2y^2}{3}-x^2yf'_u+y^2=
=-\frac{y^2}{3}-\frac{2y^2}{3}+y^2=
=y^2(-\frac{1}{3}-\frac{2}{3}+1)=
=y^2\cdot 0
= 0
We got zero as required.
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