What is the second derivative of $f (x) = \frac{8}{x} - 2$ $f(x)= 8/x-2$ ?

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Answer 1 · 2015-12-10T17:11:49

$\frac{16}{x^{3}}$ $16/(x^3)$

The first derivative is:
$\frac{d}{d x} \frac{8}{x} - \frac{d}{d x} 2$ $d/(dx)8/x - d/(dx)2$

Rewriting,
$\frac{d}{d x} 8 \cdot x^{- 1} - \frac{d}{d x} 2$ $d/(dx)8*x^-1 - d/(dx)2$

Using derivative rules, and the fact that the derivative of any constant is zero,
$8 \cdot \frac{d}{d x} x^{- 1} - 0$ $8*d/(dx)x^-1 - 0$

Using the power rule to finish,
$- 8 \cdot x^{- 2}$ $-8*x^-2$

This is our first derivative. To find the second, we simply take the derivative of the above expression:
$\frac{d}{d x} - 8 x^{- 2}$ $d/(dx)-8x^-2$

All we need to do is use the power rule again:
$16 x^{- 3}$ $16x^-3$

In another form:
$\frac{16}{x^{3}}$ $16/(x^3)$

And we're done.

What is the second derivative of f(x)= 8/x-2f(x)=8x−2f(x)= 8/x-2?