Question

How do you find the first and second derivative of `y = 2ln(x)`?

Answer 1

`y' = 2/x`
`y'' = -2/x^2`

Explanation:

We will use the product rule and the well-known derivative `(lnx)' = 1/x`.

The product rule states that for a function `f(x) = g(x) xx h(x)`, the derivative, `f'(x)`, is given by `f'(x) = g'(x) xx h(x) + h'(x) xx g(x)`.

`y' = 0 xx lnx + 2 xx 1/x`

`y' = 2/x`

We will find the second derivative using the quotient rule. The quotient rule states that for a function `f(x) = g(x)/h(x)`, the derivative is given by `f'(x) = (g'(x) xx h(x) - g(x) xx h'(x))/(h(x))^2`.

`y'' = (0 xx x - 2 xx 1)/(x)^2`

`y'' = -2/x^2`

Hopefully this helps!