What is the second derivative of the function f(x)=2x^3+e^2x+sin 2x-lnx?
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
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
giorgiana1976 | College Teacher | (Level 3) Valedictorian
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.