`f(x) = x/(x^2 - 1)` Find f'(x) and f''(x)

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Chapter 3, 3.2 - Problem 30 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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To find the derivative of the function `f(x) = x/(x^2 - 1)` use the quotient rule. The rule gives the derivative of `f(x) = (g(x))/(h(x))` as `f'(x) = (g'(x)*h(x) - g(x)*h'(x))/(h(x))^2`

`f(x) = x/(x^2-1)`

`f'(x) = (x'*(x^2 - 1) - x*(x^2 - 1)')/(x^2 - 1)^2`

= `(x^2 - 1 - 2x^2)/(x^2 - 1)^2`

= `(-x^2 - 1)/(x^2 - 1)^2`

= `-(x^2 + 1)/(x^2 - 1)^2`

`f''(x) = (-(x^2 + 1)/(x^2 - 1))'`

= `((-x^2 - 1)'*(x^2 - 1)^2 + (x^2 +1)*((x^2 - 1)^2)')/(x^2 - 1)^4`

= `(-2x*(x^2 - 1)^2 + (x^2 + 1)*2*(x^2 - 1)*2x)/(x^2 - 1)^4`

= `(-2x*(x^2 - 1) + (x^2 + 1)*2*2x)/(x^2 - 1)^3`

= `(-2x^3 + 2x + 4x^3 + 4x)/(x^2 - 1)^3`

= `(2x^3 + 6x)/(x^2 - 1)^3`

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