Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 3375

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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