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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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