How do you find the first and second derivative of (lnx)^2?

Jan 12, 2018

$\text{see explanation}$

Explanation:

$\text{differentiate using the "color(blue)"chain rule}$

$\text{given "y=f(g(x))" then}$

dy/dx=f'(g(x)xxg'(x)larrcolor(blue)"chain rule"

$\Rightarrow \frac{d}{\mathrm{dx}} \left({\left(\ln x\right)}^{2}\right)$

$= 2 \ln x \times \frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{2 \ln x}{x} \leftarrow \textcolor{b l u e}{\text{first derivative}}$

$\text{differentiate first derivative to obtain second derivative}$

$\frac{d}{\mathrm{dx}} \left(\frac{2 \ln x}{x}\right)$

$\text{differentiate using the "color(blue)"product rule}$

$\text{given "y=g(x)h(x)" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = g \left(x\right) h ' \left(x\right) + h \left(x\right) g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{product rule}}$

$\text{express } \frac{2 \ln x}{x} = 2 \ln x . {x}^{-} 1$

$g \left(x\right) = 2 \ln x \Rightarrow g ' \left(x\right) = \frac{2}{x}$

$h \left(x\right) = {x}^{-} 1 \Rightarrow h ' \left(x\right) = - {x}^{-} 2 = - \frac{1}{x} ^ 2$

$\Rightarrow \frac{d}{\mathrm{dx}} \left(2 \ln x . {x}^{-} 1\right)$

$= - 2 \ln x . \frac{1}{x} ^ 2 + \frac{1}{x} . \frac{2}{x}$

$= \frac{2}{x} ^ 2 - \frac{2 \ln x}{x} ^ 2 \leftarrow \textcolor{b l u e}{\text{second derivative}}$