what is the derivative of f(x)=(x^2)/(1-x^2)?what is the value of x when f'(x)=0?

Answer 1

f(x) = x²/(1-x²)

To take the derivative use the quotient rule.

f'(x) = {2x(1-x²) - (-2x)x²}/(1-x²)²
= {2x - 2x³ + 2x³}/(1-x²)² = 2x/(1-x²)²

f'(x) = 2x/(1-x²)² = 0
2x = 0
x = 0

So the value of x is 0 when f'(x) = 0.

Answer 2

I think the answer is 0. Anything to 0 is zero

Answer 3

(x@a)-f(x) value is x(1) @-1-2-1) + 2.9999999999