What is the second derivative of the function f(x)=2x^3+e^2x+sin 2x-lnx?

To determine the second derivative, we'll have to
determine the 1st derivative, for the beginning.

f'(x) =
(2x^3+e^2x+sin 2x-lnx)'

f'(x) = 6x^2 + 2e^2x + 2cos 2x -
1/x

Now, we'll determine the second
derivative:

f"(x) =
[f'(x)]'

In other words, we'll determine the derivative of
the expression of the 1st derivative:

f"(x) = (6x^2 + 2e^2x
+ 2cos 2x - 1/x)'

f"(x) = 12x + 4e^2x - 4sin 2x +
1/x^2

The second derivative is: f"(x) = 12x +
4e^2x - 4sin 2x + 1/x^2.