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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