# What is the second derivative of x^2 + (16/x)?

Oct 29, 2016

$\therefore {d}^{2} / {\mathrm{dx}}^{2} \left({x}^{2} + \frac{16}{x}\right) = 2 + \frac{32}{x} ^ 3$

#### Explanation:

You should learn the power rule for differentiation, which is that:
$\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1} \forall n \in \mathbb{R}$

So, $\frac{d}{\mathrm{dx}} \left({x}^{2} + \frac{16}{x}\right) = \frac{d}{\mathrm{dx}} \left({x}^{2} + 16 {x}^{-} 1\right)$
$\therefore \frac{d}{\mathrm{dx}} \left({x}^{2} + \frac{16}{x}\right) = 2 {x}^{2 - 1} + 16 \left(- 1\right) {x}^{- 1 - 1}$
$\therefore \frac{d}{\mathrm{dx}} \left({x}^{2} + \frac{16}{x}\right) = 2 x - 16 {x}^{-} 2$

And differentiating a second time, gives us:
${d}^{2} / {\mathrm{dx}}^{2} \left({x}^{2} + \frac{16}{x}\right) = 2 \left(1\right) {x}^{1 - 0} - 16 \left(- 2\right) {x}^{- 2 - 1}$
$\therefore {d}^{2} / {\mathrm{dx}}^{2} \left({x}^{2} + \frac{16}{x}\right) = 2 {x}^{0} + 32 {x}^{-} 3$
$\therefore {d}^{2} / {\mathrm{dx}}^{2} \left({x}^{2} + \frac{16}{x}\right) = 2 + \frac{32}{x} ^ 3$