Exercise
Given the differentiable function
z(x,y)=e^yf(ye^{\frac{x^2}{2y^2}})
Prove the equation
(x^2-y^2)\cdot z'_x+xy\cdot z'_y=xyz
Proof
When we have a function and in its parentheses there is a complex expression instead of a simple variable, we will define a new variable like this
t=ye^{\frac{x^2}{2y^2}}
We get the function
z(x,y)=e^yf(t)
Or simply
z(x,y)=e^yf
We will use the chain rule to calculate the partial derivatives of z.
z'_x=e^y\cdot f'_t\cdot t'_x
z'_y=e^y\cdot f + e^y\cdot f'_t\cdot t'_y
Similarly, we will calculate the partial derivatives of t.
t'_x=ye^{\frac{x^2}{2y^2}}\cdot \frac{2x}{2y^2}=
=ye^{\frac{x^2}{2y^2}}\cdot \frac{x}{y^2}=
=e^{\frac{x^2}{2y^2}}\cdot \frac{x}{y}
t'_y=1\cdot e^{\frac{x^2}{2y^2}}+y\cdot e^{\frac{x^2}{2y^2}}\cdot \frac{-x^2}{4y^4}\cdot 4y=
t'_y=e^{\frac{x^2}{2y^2}}(1-y\cdot \frac{x^2}{y^3})=
t'_y=e^{\frac{x^2}{2y^2}}(1-\frac{x^2}{y^2})
We put the results in the partial derivatives of z and get
z'_x=e^y\cdot f'_t\cdot e^{\frac{x^2}{2y^2}}\cdot \frac{x}{y}
z'_y=e^y\cdot f + e^y\cdot f'_t\cdot e^{\frac{x^2}{2y^2}}(1-\frac{x^2}{y^2})
We will put the partial derivatives in the left side of the equation we need to prove.
(x^2-y^2)\cdot z'_x+xy\cdot z'_y=
=(x^2-y^2)\cdot e^y\cdot f'_t\cdot e^{\frac{x^2}{2y^2}}\cdot \frac{x}{y}+xy\cdot (e^y\cdot f + e^y\cdot f'_t\cdot e^{\frac{x^2}{2y^2}}(1-\frac{x^2}{y^2})=
=\frac{x^3}{y}e^ye^{\frac{x^2}{2y^2}}f'_t-xye^ye^{\frac{x^2}{2y^2}}f'_t+xye^yf +xye^yf'_t e^{\frac{x^2}{2y^2}}-xye^yf'_t\frac{x^2}{y^2}e^{\frac{x^2}{2y^2}}=
=\frac{x^3}{y}e^ye^{\frac{x^2}{2y^2}}f'_t+xye^yf +-xye^yf'_t\frac{x^2}{y^2}e^{\frac{x^2}{2y^2}}=
=\frac{x^3}{y}e^ye^{\frac{x^2}{2y^2}}f'_t+xye^yf +-e^yf'_t\frac{x^3}{y}e^{\frac{x^2}{2y^2}}=
=xye^yf=
We put the equation
z=e^yf
And get
=xyz
Hence, we got
(x^2-y^2)\cdot z'_x+xy\cdot z'_y=xyz
As required.
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