What is the first derivative and second derivative of #(2x^2)/(x^2-1)#?

Question

Cameron Alexander · Accepted Answer

#dy/dx = -(4x)/(x^2 - 1)^2 #

#(d^2y)/dx^2 = (12x^2+4)/(x^2-1)^3# Given: #(2x^2)/(x^2 - 1)# Add 0 to the numerator in the form of #- 2 + 2#: #(2x^2 - 2 + 2)/(x^2 - 1) # Split into 2 fractions: #(2x^2 - 2)/(x^2 - 1) + 2/(x^2 - 1) # The first fraction reduces to 2 and the second fraction can be written as a negative power: #2 + 2(x^2 - 1)^-1 # The first derivative is: #dy/dx = -4x(x^2 - 1)^-2 # #dy/dx = -(4x)/(x^2 - 1)^2 # #(d^2y)/dx^2 = (12x^2+4)/(x^2-1)^3#

Adam Adkins · Answer

undefined To find the first derivative of the function (2x^2)/(x^2-1), apply the quotient rule. The first derivative, f'(x), is:

f'(x) = [(2x^2)'(x^2-1) - (2x^2)(x^2-1)'] / (x^2-1)^2

Now, differentiate each term:
f'(x) = [(4x)(x^2-1) - (2x^2)(2x)] / (x^2-1)^2
f'(x) = [4x^3 - 4x - 4x^3] / (x^2-1)^2
f'(x) = (-4x) / (x^2-1)^2

To find the second derivative, differentiate f'(x) with respect to x:
f''(x) = (-4(x^2-1)^2 - (-4x)(2(x^2-1)(2x))) / (x^2-1)^4
f''(x) = (-4(x^4 - 2x^2 + 1) + 8x(x^2-1)(2x)) / (x^2-1)^4
f''(x) = (-4x^4 + 8x^2 - 4 - 16x^4 + 16x^2) / (x^2-1)^4
f''(x) = (-20x^4 + 24x^2 - 4) / (x^2-1)^4