# Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 3370

Exercise

Given the differentiable function

$$z(x,y)=f(x^2-y^2)$$

Prove the equation

$$y\cdot z'_x+x\cdot z'_y=0$$

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

$$t=x^2-y^2$$

We will use the chain rule to calculate the partial derivatives of z.

$$z'_x=f'_t\cdot t'_x$$

$$z'_y=f'_t\cdot t'_y$$

Also,we will  calculate the partial derivatives of t.

$$t'_x=2x$$

$$t'_y=-2y$$

We will put the results in the derivative of z

$$z'_x=f'_t\cdot 2x$$

$$z'_y=f'_t\cdot (-2y)$$

We will put the partial derivatives in the left side of the equation we need to prove.

$$y\cdot z'_x+x\cdot z'_y=$$

$$=y\cdot f'_t\cdot 2x-x\cdot f'_t\cdot 2y=$$

$$=2xy\cdot f'_t-2xy\cdot f'_t=0$$

Hence, we got

$$y\cdot z'_x+x\cdot z'_y=0$$

As required.

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