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

Oct 23, 2017

f’(x) = - (x^2 + 4 ) / (x^2 - 4)^2

f”(x) = (2x*(x^2 +12))/(x^2-4)^3

#### Explanation:

$f \left(x\right) = \frac{x}{{x}^{2} - 4}$

$u = x , \mathrm{du} = 1$

$v = \left({x}^{2} - 4\right) , \mathrm{dv} = 2 x$

Quotient rule $\frac{v . \mathrm{du} - u . \mathrm{dv}}{v} ^ 2$

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.
$u = - \left({x}^{2} + 4\right) , \mathrm{du} = - 2 x$

$v = {\left({x}^{2} - 4\right)}^{2} , \mathrm{dv} = 2 \left({x}^{2} - 4\right) \cdot 2 x = 4 x \left({x}^{2} - 4\right)$

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#