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Hollysmoke
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I just wanted to make sure I differentiated this correctly before I started graphing it:
y=x^2/1-x^2
y' = (2x)/(1-x^2)^2
y''=(2+6x^2)/(1-x^2)^3
y=x^2/1-x^2
y' = (2x)/(1-x^2)^2
y''=(2+6x^2)/(1-x^2)^3
Hollysmoke said:I just wanted to make sure I differentiated this correctly before I started graphing it:
y=x^2/1-x^2
y' = (2x)/(1-x^2)^2
y''=(2+6x^2)/(1-x^2)^3
The equation for differentiating y = x^2/1-x^2 is y' = (2x(1-x^2) - 2x^3)/(1-x^2)^2.
The process for differentiating y = x^2/1-x^2 is to use the quotient rule, which states that the derivative of a quotient is equal to (the derivative of the numerator times the denominator) minus (the derivative of the denominator times the numerator), all divided by the square of the denominator.
Yes, the equation y' = (2x(1-x^2) - 2x^3)/(1-x^2)^2 can be simplified to y' = 2x/(1-x^2) - 2x^3/(1-x^2)^2.
The domain of y = x^2/1-x^2 is all real numbers except for x = 1 and x = -1, as these values would make the denominator equal to 0.
The graph of y = x^2/1-x^2 is a hyperbola with a vertical asymptote at x = 1 and a horizontal asymptote at y = 1. The graph approaches the asymptotes but does not touch them.