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

To find the first and second derivatives of $\frac{\ln(x^2)}{x}$, you can use the quotient rule and the chain rule. Let's denote $u = \ln(x^2)$ and $v = x$. Then, applying the quotient rule:

1. Find the first derivative: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$

2. Find the second derivative: $\frac{d^2}{dx^2}\left(\frac{u}{v}\right) = \frac{v \frac{d^2u}{dx^2} - 2\frac{du}{dx} \frac{dv}{dx} - u \frac{d^2v}{dx^2}}{v^2}$

Now, let's compute the derivatives:

1. First derivative: $\frac{du}{dx} = \frac{1}{x^2}(2x) = \frac{2}{x}$ $\frac{dv}{dx} = 1$

$\frac{d}{dx}\left(\frac{\ln(x^2)}{x}\right) = \frac{x(2/x) - \ln(x^2)(1)}{x^2} = \frac{2 - 2\ln(x)}{x^2}$

2. Second derivative: $\frac{d^2u}{dx^2} = \frac{d}{dx}\left(\frac{2}{x}\right) = -\frac{2}{x^2}$ $\frac{d^2v}{dx^2} = 0$

$\frac{d^2}{dx^2}\left(\frac{\ln(x^2)}{x}\right) = \frac{x(-2/x^2) - 2\left(\frac{2}{x}\right)(1) - \ln(x^2)(0)}{x^2}$ $= \frac{-2 + 4/x^2}{x^2} = \frac{-2x^2 + 4}{x^4}$

So, the first derivative is $\frac{2 - 2\ln(x)}{x^2}$ and the second derivative is $\frac{-2x^2 + 4}{x^4}$.