# How do you find the first and second derivative of ln (x^8)/ x^2?

Sep 25, 2017

$\left(1\right) : \text{ The First Derivative is, } \frac{8}{x} ^ 3 \left({x}^{2} - 2 \ln x\right) .$

$\left(2\right) : \text{ The Second Derivative is, } \frac{24}{x} ^ 4 \left(2 \ln x - {x}^{2}\right) .$

#### Explanation:

Let, $f \left(x\right) = \ln \frac{{x}^{8}}{x} ^ 2 = \frac{8 \ln x}{x} ^ 2. \ldots . \left[\because , \ln {a}^{m} = m \ln a\right] ,$

$\therefore f \left(x\right) = 8 \cdot {x}^{-} 2 \cdot \ln x .$

By the Product Rule, then, we get,

$f ' \left(x\right) = 8 \left\{{x}^{-} 2 \frac{d}{\mathrm{dx}} \left(\ln x\right) + \left(\ln x\right) \frac{d}{\mathrm{dx}} \left({x}^{-} 2\right)\right\} ,$

$= 8 \left\{{x}^{-} 2 \cdot \frac{1}{x} + \left(- 2 \cdot {x}^{- 2 - 1}\right) \ln x\right\} ,$

$= 8 \left\{{x}^{-} 1 - 2 {x}^{-} 3 \cdot \ln x\right\} ,$

$= 8 \left\{\frac{1}{x} - \frac{2 \ln x}{x} ^ 3\right\} ,$

$\Rightarrow f ' \left(x\right) = \frac{8}{x} ^ 3 \left({x}^{2} - 2 \ln x\right) .$

Knowing that, $f ' ' \left(x\right) = \left\{f ' \left(x\right)\right\} ' ,$ we have,

$f ' ' \left(x\right) = \left\{8 \left({x}^{-} 1 - 2 {x}^{-} 3 \cdot \ln x\right)\right\} ' ,$

$= 8 \left({x}^{-} 1 - 2 {x}^{-} 3 \cdot \ln x\right) ' ,$

$= 8 \left({x}^{-} 1\right) ' - 8 \cdot 2 \left({x}^{-} 3 \cdot \ln x\right) ' ,$

$= 8 \left(- 1 \cdot {x}^{- 1 - 1}\right) - 16 \left\{{x}^{-} 3 \frac{d}{\mathrm{dx}} \left(\ln x\right) + \left(\ln x\right) \frac{d}{\mathrm{dx}} \left({x}^{-} 3\right)\right\} ,$

$= - 8 {x}^{-} 2 - 16 \left\{{x}^{-} 3 \cdot \frac{1}{x} + \left(- 3 \cdot {x}^{- 3 - 1}\right) \ln x\right\} ,$

$= - \frac{8}{x} ^ 2 - 16 \left(\frac{1}{x} ^ 2 - \frac{3 \ln x}{x} ^ 4\right) ,$

$= - \frac{8}{x} ^ 2 - \frac{16}{x} ^ 2 + \frac{48 \ln x}{x} ^ 4 ,$

$= - \frac{24}{x} ^ 2 + \frac{48 \ln x}{x} ^ 4.$

$\Rightarrow f ' ' \left(x\right) = \frac{24}{x} ^ 4 \left(2 \ln x - {x}^{2}\right) .$

Enjoy Maths.!