Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6509

Exercise

Given the differentiable function

$u=\frac{xy}{z}\ln x+xf(\frac{y}{x},\frac{z}{x})$

Prove the equation

$xu'_x+yu'_y+zu'_z=u+\frac{xy}{z}$

Proof

Define

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

$w=\frac{z}{x}$

We get the function

$u=\frac{xy}{z}\ln x+xf(v,w)=\frac{xy}{z}\ln x+xf$

And the internal functions

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

$w(x,z)=\frac{z}{x}$

We will use the chain rule to calculate the partial derivatives of u.

$u'_x=\frac{y}{z}\ln x+\frac{xy}{z}\cdot\frac{1}{x}+1\cdot f+x\cdot f'_x=$

$=\frac{y}{z}(\ln x+1)+f+x(f'_v\cdot v'_x+f'_w\cdot w'_x)=$

$=\frac{y}{z}(\ln x+1)+f+x(f'_v\cdot (-\frac{y}{x^2})+f'_w\cdot (-\frac{z}{x^2}))=$

$=\frac{y}{z}(\ln x+1)+f-\frac{y}{x}f'_v-\frac{z}{x}f'_w$

$u'_y=\frac{x}{z}\ln x+x\cdot f'_y=$

$=\frac{x\ln x}{z}+x(f'_v\cdot v'_y+f'_w\cdot w'_y)=$

$=\frac{x\ln x}{z}+xf'_v\cdot \frac{1}{x}+0=$

$=\frac{x\ln x}{z}+f'_v$

$u'_z=-\frac{xy}{z^2}\ln x+x\cdot f'_z=$

$=-\frac{xy}{z^2}\ln x+x(f'_v\cdot v'_z+f'_w\cdot w'_z)=$

$=-\frac{xy}{z^2}\ln x+0+xf'_w\cdot\frac{1}{x}=$

$=-\frac{xy}{z^2}\ln x+f'_w$

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

$xu'_x+yu'_y+zu'_z=$

$=x(\frac{y}{z}(\ln x+1)+f-\frac{y}{x}f'_v-\frac{z}{x}f'_w)+y(\frac{x\ln x}{z}+f'_v)+z(-\frac{xy}{z^2}\ln x+f'_w)=$

$=\frac{xy}{z}(\ln x+1)+xf-yf'_v-zf'_w+\frac{xy\ln x}{z}+yf'_v-\frac{xy}{z}\ln x+zf'_w=$

$=\frac{xy}{z}(\ln x+1)+xf+\frac{xy\ln x}{z}-\frac{xy}{z}\ln x=$

$=\frac{xy}{z}\ln x+\frac{xy}{z}+xf+\frac{xy\ln x}{z}-\frac{xy}{z}\ln x=$

$=\frac{xy}{z}+xf+\frac{xy\ln x}{z}=$

$=\frac{xy}{z}+u$

We got to the right side of the equation as required.

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