# 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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