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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