Question

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

Answer 1

The First Derivative is:

`dy/dx = 1/(xlnx)`

The Second Derivative is:

`(d^2y)/(dx^2) = - (1 + lnx)/(x^2ln^2x)`

Explanation:

We have:

`y = ln(lnx^2)`

Using the law of logarithms we can write this as:

`y = ln(2lnx)`
`\ \ = ln2 + ln(lnx)`

Then Differentiating wrt `x` and applying the chain rule:

`dy/dx = 0 + 1/lnx * 1/x`
`" " = 1/(xlnx)`

If we write this as:

`dy/dx = (xlnx)^(-1)`

Then we get the second derivative by differentiating again wrt `x` and applying the chain and product rule:

`(d^2y)/(dx^2) = (-1)(xlnx)^(-2)(x 1/x + 1.lnx)`
`" " = -1/(xlnx)^2 \ (1 + lnx)`
`" " = - (1 + lnx)/(x^2ln^2x)`