# Question 79c72

Jan 8, 2018

$2 {x}^{-} 3$ or $\frac{2}{x} ^ 3$

#### Explanation:

$y = \frac{1}{x}$
$y = {x}^{- 1}$

power rule: $y ' = n {x}^{n - 1}$, where $n = - 1$

$y ' = - 1 \cdot {x}^{-} 2 = - 1 {x}^{-} 2$ or $- \frac{1}{x} ^ 2$

$y ' ' = \frac{\delta y}{\delta x} \left(- {x}^{-} 2\right)$

$y ' ' = n {x}^{n - 1}$, where $n = - 2$
$y ' ' = - 2 \cdot - {x}^{- 3} = 2 {x}^{-} 3$ or $\frac{2}{x} ^ 3$

Jan 8, 2018

Differentiate twice.
Answer $2 {x}^{-} 3$

#### Explanation:

First we need to simplify the expression we are truing to differentiate.

From Laws of indices, we know that $\frac{1}{x} = {x}^{-} 1$

Now to differentiate, we multiply the coefficient by the power, and then $- 1$ from the power.

$\frac{\mathrm{dy}}{\mathrm{dx}} = n a {x}^{n} - 1$ where a is the coefficient and n# is the power.

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} {x}^{-} 1 = - 1 {x}^{-} 2 = - {x}^{-} 2$

However, we are finding the second derivative so in turn we must differentiate once more.

$\therefore \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 2 {x}^{-} 3$