# 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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