# 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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