Find y' and y''
y=X^3/(X^2-1)
you will have to use the quotient rule twice
y' = [(x^2-1)(3x^2) - x^3(2x)]/(x^2-1)^2
= (x^4 - 3x^2)/(x^2 - 1)^2
then
y'' = [(x^2-1)^2(4x^3-6x) - (x^4-3x^2)(2)(x^2-1)(2x)]/(x^2-1)^4
I then took out a common factor of (x^2-1), simplified the rest to get
y'' = 2x(x^2+3)/(x^2-1)^3
I will let you try the algebra, let me know if you got something different.
I got the same answer for the first derivative but I am still messing something up for the second derivative. It is probably taking out the common factor. I'll check back later, got to go.
To find the first derivative (y') of y = X^3 / (X^2 - 1), you can use the quotient rule.
The quotient rule states that for the function f(x) = g(x) / h(x), the derivative f'(x) is given by:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
Now, let's apply the quotient rule to find y':
First, we need to find g'(x) and h'(x).
g(x) = X^3
g'(x) = 3X^2
h(x) = X^2 - 1
h'(x) = 2X
Now, substitute these values into the quotient rule formula:
y' = (3X^2 * (X^2 - 1) - X^3 * 2X) / ((X^2 - 1)^2)
= (3X^4 - 3X^2 - 2X^4) / ((X^2 - 1)^2)
= (X^4 - 3X^2) / (X^2 - 1)^2
Therefore, y' = (X^4 - 3X^2) / (X^2 - 1)^2
To find the second derivative (y''), you'll differentiate y' with respect to X.
Differentiating y' = (X^4 - 3X^2) / (X^2 - 1)^2, we can use the quotient rule again:
Applying the quotient rule:
y'' = [(4X^3 * (X^2 - 1)^2 - 2(X^4 - 3X^2) * 2X * (X^2 - 1)) / (X^2 - 1)^4]
/ [(X^2 - 1)^2]
Simplifying further will give you the final expression for y'':
y'' = [(4X^3 * (X^2 - 1)^2 - 4X^3 * (X^2 - 1)) / (X^2 - 1)^4]
/ [(X^2 - 1)^2]
Therefore, y'' = (4X^3 * (X^2 - 1) - 4X^3) / (X^2 - 1)^3