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

### 2 Answers | Add Yours

We have the function f(x) =2x^3 + e^2x + sin 2x - ln x. We have to find f''(x).

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

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

**The required second derivative f(x) = 12x + 4e^2x - 4 sin 2x + 1/x^2**

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.**