Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6801

Exercise

Given the differentiable function

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

Prove the equation

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

Proof

Define

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

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

We get the function

$u=f(v,w)+\frac{xy}{z}$

And the internal functions

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

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

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

$u'_x=f'_v\cdot v'_x+f'_w\cdot w'_x+\frac{y}{z}=$

$=f'_v\cdot (-\frac{y}{x^2})+f'_w\cdot 0+\frac{y}{z}=$

$=-\frac{y}{x^2}f'_v+\frac{y}{z}$

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

$=f'_v\cdot \frac{1}{x}+f'_w\cdot \frac{1}{z}+\frac{x}{z}=$

$=\frac{1}{x}f'_v + \frac{1}{z}f'_w+\frac{x}{z}$

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

$=f'_v\cdot 0+f'_w\cdot (-\frac{y}{z^2})-\frac{xy}{z^2}=$

$=-\frac{y}{z^2}f'_w-\frac{xy}{z^2}$

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

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

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

$=-\frac{y}{x}f'_v+\frac{xy}{z}+\frac{y}{x}f'_v+\frac{y}{z}f'_w+\frac{xy}{z}-\frac{y}{z}f'_w-\frac{xy}{z}=$

$=\frac{xy}{z}$

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

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