Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6522

Exercise

Given the differentiable function

$z=xf(\frac{y}{x})+g(\frac{y}{x})$

Prove the equation

$x^2z''_{xx}+2xyz''_{xy}+y^2z''_{yy}=0$

Proof

Define

$u=\frac{y}{x}$

We get the function

$z=xf(u)+g(u)$

And the internal function

$u(x,y)=\frac{y}{x}$

We will use the chain rule to calculate the first and second order partial derivatives of z.

$z'_x=1\cdot f+x\cdot f'_u\cdot u'_x+g'_u\cdot u'_x=$

$=f+x\cdot f'_u\cdot (-\frac{y}{x^2})+g'_u\cdot (-\frac{y}{x^2})=$

$=f-\frac{y}{x}\cdot f'_u-\frac{y}{x^2}\cdot g'_u$

$z''_{xx}=f'_u\cdot u'_x+f'_u+x\cdot f''_u\cdot u'_x+y\cdot g''_u\cdot u'_x=$

$=f'_u\cdot u'_x-(-\frac{y}{x^2}\cdot f'_u+\frac{y}{x}\cdot f''_u\cdot u'_x)-(-\frac{y}{x^2}\cdot 2x\cdot g'_u+\frac{y}{x^2}\cdot g''_u\cdot u'_x)=$

$=-\frac{y}{x^2}\cdot f'_u+\frac{y}{x^2}\cdot f'_u+\frac{y}{x}\cdot\frac{y}{x^2}\cdot f''_u+\frac{2xy}{x^4}\cdot g'_u+\frac{y^2}{x^4}\cdot g''_u$

$=\frac{y}{x}\cdot\frac{y}{x^2}\cdot f''_u+\frac{2xy}{x^4}\cdot g'_u+\frac{y^2}{x^4}\cdot g''_u$

$z''_{xy}=f'_u\cdot u'_y-(\frac{1}{x}\cdot f'_u+\frac{y}{x}\cdot f''_u\cdot u'_y)-(\frac{1}{x^2}\cdot g'_u+\frac{y}{x^2}\cdot g''_u\cdot u'_y)=$

$=\frac{1}{x}\cdot f'_u-\frac{1}{x}\cdot f'_u-\frac{y}{x^2}\cdot f''_u-\frac{1}{x^2}\cdot g'_u-\frac{y}{x^3}\cdot g''_u=$

$=-\frac{y}{x^2}\cdot f''_u-\frac{1}{x^2}\cdot g'_u-\frac{y}{x^3}\cdot g''_u$

$z'_y=x\cdot f'_u\cdot u'_y+g'_u\cdot u'_y=$

$=x\cdot f'_u\cdot \frac{1}{x}+g'_u\cdot\frac{1}{x}=$

$=f'_u+\frac{1}{x}\cdot g'_u$

$z''_{yy}=f''_u\cdot u'_y+\frac{1}{x}\cdot g''_u\cdot u'_y=$

$=\frac{1}{x}\cdot f''_u+\frac{1}{x^2}\cdot g''_u$

We will put the partial derivatives in the left side of the equation we need to prove.

$x^2z''_{xx}+2xyz''_{xy}+y^2z''_{yy}=$

$=x^2(\frac{y}{x}\cdot\frac{y}{x^2}\cdot f''_u+\frac{2xy}{x^4}\cdot g'_u+\frac{y^2}{x^4}\cdot g''_u)+2xy(-\frac{y}{x^2}\cdot f''_u-\frac{1}{x^2}\cdot g'_u-\frac{y}{x^3}\cdot g''_u)+y^2(\frac{1}{x}\cdot f''_u+\frac{1}{x^2}\cdot g''_u)=$

$=\frac{y^2}{x}\cdot f''_u+\frac{2y}{x}\cdot g'_u+\frac{y^2}{x^2}\cdot g''_u-\frac{2y^2}{x}\cdot f''_u-\frac{2y}{x}\cdot g'_u-\frac{2y^2}{x^2}\cdot g''_u+\frac{y^2}{x}\cdot f''_u+\frac{y^2}{x^2}\cdot g''_u=$

$=0$

We got zero as required.

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