Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6506

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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