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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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