# How do you find the first and second derivative of (x^3-3x^2+8x+18)/x?

Jun 12, 2018

Shown below

#### Explanation:

$y = \frac{{x}^{3} - 3 {x}^{2} + 8 x + 18}{x}$

$y = {x}^{3} / x - \frac{3 {x}^{2}}{x} + \frac{8 x}{x} + \frac{18}{x}$

$y = {x}^{2} - 3 x + 8 + 18 {x}^{- 1}$

$\implies \frac{d}{\mathrm{dx}} \left(\frac{{x}^{3} - 3 {x}^{2} + 8 x + 18}{x}\right) = \frac{d}{\mathrm{dx}} \left({x}^{2} - 3 x + 8 + 18 {x}^{- 1}\right)$

Use the power rule:

color(red)(d/dx ( x^n ) = nx^(n-1)

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x - 3 - 18 {x}^{- 2}$

$\implies \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{d}{\mathrm{dx}} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

$\implies \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{d}{\mathrm{dx}} \left(2 x - 3 - 18 {x}^{- 2}\right)$

$\implies \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 2 + 36 {x}^{- 3}$

Jun 12, 2018

$\left(i\right) : f ' \left(x\right) = 2 x - 3 - \frac{18}{x} ^ 2 = \frac{2 {x}^{3} - 3 x - 18}{x} ^ 2$.

$\left(i i\right) : f ' ' \left(x\right) = 2 + \frac{36}{x} ^ 3 = \frac{2 \left({x}^{3} + 18\right)}{x} ^ 3$.

#### Explanation:

Prerequisite : $\left({x}^{n}\right) ' = n {x}^{n - 1}$.

Let $f \left(x\right) = \frac{{x}^{3} - 3 {x}^{2} + 8 x + 18}{x}$.

$\therefore f \left(x\right) = {x}^{3} / x - \frac{3 {x}^{2}}{x} + \frac{8 x}{x} + \frac{18}{x} , i . e . ,$

$f \left(x\right) = {x}^{2} - 3 x + 8 + \frac{18}{x}$.

$\therefore f ' \left(x\right) = \left({x}^{2}\right) ' - 3 \left(x\right) ' + 0 + 18 \left({x}^{-} 1\right) '$,

$= 2 {x}^{2 - 1} - 3 \left(1 {x}^{1 - 1}\right) + 18 \left(- 1 {x}^{- 1 - 1}\right)$.

$\Rightarrow f ' \left(x\right) = 2 x - 3 - \frac{18}{x} ^ 2 = \frac{2 {x}^{3} - 3 x - 18}{x} ^ 2$.

Similarly, $f ' ' \left(x\right) = \left(f ' \left(x\right)\right) '$,

$= \left(2 x - 3 - 18 {x}^{-} 2\right) '$,

$= 2 - 0 - 18 \left(- 2 {x}^{- 2 - 1}\right)$.

$\Rightarrow f ' ' \left(x\right) = 2 + \frac{36}{x} ^ 3 = \frac{2 \left({x}^{3} + 18\right)}{x} ^ 3$.