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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