Question

How do you find the first and second derivative of `{ln(x)}^2`?

Answer 1

Let your function be `f(x) = (lnx)^2`, then by the chain rule, we have:

`f'(x) = 1/x xx 2lnx = (2ln(x))/x`

This is our first derivative. We obtain the second derivative by differentiating the first derivative.

If `f'(x) = g(x)/(h(x))`, then `f''(x) = (g'(x) xx h(x) - g(x) xx h'(x))/(h(x))^2`

`g(x) = 2lnx -> g'(x) = 2/x`

`h(x) = x -> h'(x) = 1`

`f''(x) = (2/x xx x - 2lnx xx 1)/(x)^2`

`f''(x) = (2 - 2lnx)/x^2 "or" (2(1 - lnx))/x^2`

In summary:

The first derivative of `{ln(x)}^2` is `dy/dx = (2ln(x))/x`

The second derivative of `{ln(x)}^2` is `(d^2y)/(dx^2) = (2(1 - lnx))/x^2`.

Practice exercises:

1. Determine the first, and second derivatives, if they exist.

a) `y = (x - 2)^(-1)`

b) `y = 5x^5 - 7x^4 + 3x^3 - 9x^2 + 2`

c) `f(x) = e^(15x) xx lnx^3`

d) `(ln(sin^2x))^3`

Bonus:

Determine the third derivatives of the functions above, if they exist.

Good luck, and hopefully this helps!