Second derivative

Question

f'(x)= -2x/ (x^2-1)^2 

 

Hi, I keep messing the algebra for this derivative. Please show me steps. 

Answer 1

The second derivative of f'(x)

 

factor out constant

 

-2* d/dx (x/(x²-1)²

 

Use the quotient rule:

d/dx [ h(x)/g(x) ] = [h'(x)*g(x) - g'(x)*h(x) ] / g(x)²

 

Note that the derivative of the function in the denominator, g'(x) is chain ruled

g(x) = (x²-1)²

g'(x) = 2(x²-1)*(2x) = 4x(x²-1)

 

Then we have;

 

f''(x) = -2 * [ (x²-1)² - 4x(x²-1)(x) ] / (x²-1)⁴

 

factor out an (x²-1) and combine like terms

 

f''(x) = -2 * (x² - 1 - 4x²) / (x²-1)³

 

f''(x) = -2 * (-3x² - 1) / (x²-1)³

 

f''(x) = (6x²+2)/(x²-1)³