What is the second derivative of (x^3)/(x^2-1)?
To find the second derivative of a function, we need to differentiate it twice.
Step 1: Let's start by finding the first derivative of the function using the quotient rule.
The quotient rule states that if you have a function of the form f(x) = g(x) / h(x), then the derivative of f(x) with respect to x is given by (h(x)*g'(x) - g(x)*h'(x)) / [h(x)]^2.
In our case, f(x) = (x^3) / (x^2 - 1), so we can apply the quotient rule.
Let g(x) = x^3 and h(x) = x^2 - 1.
g'(x) = 3x^2 (using the power rule for differentiation)
h'(x) = 2x (using the power rule for differentiation)
Now we can apply the quotient rule:
f'(x) = [(x^2 - 1)*(3x^2) - (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]
Step 2: Now, let's find the second derivative. We will differentiate the first derivative obtained in step 1 with respect to x.
To differentiate (x^4 - 3x^2) / [(x^2 - 1)^2], we can use the quotient rule again.
Let p(x) = x^4 - 3x^2 and q(x) = (x^2 - 1)^2.
p'(x) = 4x^3 - 6x (using the power rule for differentiation)
q'(x) = 2(x^2 - 1)(2x) = 4x(x^2 - 1) (using the power rule for differentiation)
Now applying the quotient rule:
f''(x) = [(x^2 - 1)^2*(4x^3 - 6x) - (x^4 - 3x^2)*(4x(x^2 - 1))] / [(x^2 - 1)^4]
= [4x^5 - 4x - 14x^3 + 12x^3] / [(x^2 - 1)^3]
= (4x^5 - 2x^3 - 4x) / [(x^2 - 1)^3]
So, the second derivative of (x^3) / (x^2 - 1) is (4x^5 - 2x^3 - 4x) / [(x^2 - 1)^3].