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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