# What are the first and second derivatives of f(x)=-3x^5lnx ?

Dec 24, 2015

Here, we'll need the product rule, which states that for $y = f \left(x\right) g \left(x\right)$, $y ' = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

#### Explanation:

Solving:

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \left(- 15 {x}^{4}\right) \left(\ln x\right) + \left(- 3 {x}^{5}\right) \left(\frac{1}{x}\right) = - 15 {x}^{4} \ln x - 3 {x}^{4} = \textcolor{g r e e n}{- 3 {x}^{4} \left(5 \ln x + 1\right)}$

Second derivative follows the same logic:

${\left(\mathrm{df} \left(x\right)\right)}^{2} / \left(\mathrm{dx}\right) = \left(- 12 {x}^{3}\right) \left(5 \ln x + 1\right) + \left(- 3 {x}^{4}\right) \left(\frac{5}{x}\right) = - 60 {x}^{3} \ln x - 12 - 15 {x}^{3}$

${\left(\mathrm{df} \left(x\right)\right)}^{2} / \left(\mathrm{dx}\right) = \textcolor{b l u e}{- 15 {x}^{3} \left(4 \ln x + 1\right) - 12}$