How do you find the first and second derivative of  lnx^2/x?

Oct 25, 2016

$\left(\frac{\ln {x}^{2}}{x}\right) ' ' = \frac{2 \left(- 3 + \ln {x}^{2}\right)}{x} ^ 3$

Explanation:

The derivative of the quotient $\frac{u \left(x\right)}{v \left(x\right)}$ is determined using Quotient Rule

color(blue)(((U(x))/(V(x)))'=(u'(x)v(x)-v'(x)u(x))/v^2

color(red)((ln(u(x)))'=((u(x))')/(u(x))

$\left(\frac{\ln {x}^{2}}{x}\right) ' = \textcolor{b l u e}{\frac{\left(\ln {x}^{2}\right) ' \cdot x - x ' \left(\ln {x}^{2}\right)}{x} ^ 2}$

$\left(\frac{\ln {x}^{2}}{x}\right) ' = \frac{\textcolor{red}{\frac{2 x}{x} ^ 2} \cdot x - 1 \left(\ln {x}^{2}\right)}{x} ^ 2$

$\left(\frac{\ln {x}^{2}}{x}\right) ' = \frac{\frac{2 {x}^{2}}{x} ^ 2 - 1 \left(\ln {x}^{2}\right)}{x} ^ 2$
$\textcolor{p u r p \le}{\left(\frac{\ln {x}^{2}}{x}\right) ' = \frac{2 - \ln {x}^{2}}{x} ^ 2}$

$\left(\frac{\ln {x}^{2}}{x}\right) ' ' = \left(\left(\frac{\textcolor{p u r p \le}{\ln {x}^{2}}}{x}\right) '\right) '$

((lnx^2)/x)''=((color(purple)((lnx^2)/x)')'
((lnx^2)/x)''=((2-lnx^2)/x^2))'

$\left(\frac{\ln {x}^{2}}{x}\right) ' ' = \textcolor{b l u e}{\frac{\left(2 - \ln {x}^{2}\right) ' {x}^{2} - \left({x}^{2}\right) ' \left(2 - \ln {x}^{2}\right)}{x} ^ 4}$
$\left(\frac{\ln {x}^{2}}{x}\right) ' ' = \frac{\left(0 - \frac{2 x}{x} ^ 2\right) {x}^{2} - \left(2 x\right) \left(2 - \ln {x}^{2}\right)}{x} ^ 4$
$\left(\frac{\ln {x}^{2}}{x}\right) ' ' = \frac{\left(\frac{- 2 x}{\cancel{x}} ^ 2\right) {\cancel{x}}^{2} - \left(2 x\right) \left(2 - \ln {x}^{2}\right)}{x} ^ 4$
$\left(\frac{\ln {x}^{2}}{x}\right) ' ' = \frac{- 2 x - \left(2 x\right) \left(2 - \ln {x}^{2}\right)}{x} ^ 4$
$\left(\frac{\ln {x}^{2}}{x}\right) ' ' = \frac{- 2 x - 4 x + 2 x \ln {x}^{2}}{x} ^ 4$
$\left(\frac{\ln {x}^{2}}{x}\right) ' ' = \frac{- 6 x + 2 x \ln {x}^{2}}{x} ^ 4$
$\left(\frac{\ln {x}^{2}}{x}\right) ' ' = 2 x \frac{- 3 + \ln {x}^{2}}{x} ^ 4$
$\left(\frac{\ln {x}^{2}}{x}\right) ' ' = \frac{2 \left(- 3 + \ln {x}^{2}\right)}{x} ^ 3$