What is the second derivative of $x/(x^2-4)$ ?

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Answer 1 · 2017-10-23T20:43:16

$f’(x) = - (x^2 + 4 ) / (x^2 - 4)^2$

$f”(x) = (2x*(x^2 +12))/(x^2-4)^3$

$f(x) = x / (x^2-4)$

$u = x, du = 1$

$v = (x^2 - 4), dv = 2x$

Quotient rule $(v.du - u.dv) / v^2$

$f’(x) = (((x^2-4).1) - (x*2x)) / (x^2 - 4)^2$

$f’(x) = (x^2 - 4 - 2x^2) / (x^2 - 4)^2$

$f’(x) = - (x^2 + 4 ) / (x^2 - 4)^2$

Similarly, the second derivative is to be done.
$u = -(x^2+4), du = -2x$

$v = (x^2 - 4)^2, dv = 2(x^2-4)*2x = 4x(x^2-4)$

$f”(x) = (((x^2-4)^2* (-2x)) +((x^2+4)* (4x*(x^2-4)))) / (x^2-4)^4$
Taking out #(x^2-4) and simplifying,

$f”(x) = (cancel(x^2-4) (-2x^3 + 8x + 4x^3 + 16x))/(cancel(x^2-4)(x^2-4)^3$

$f”(x) = (2x*(x^2 + 12))/(x^2-4^3$

What is the second derivative of x/(x^2-4)x/(x^2-4)?