What is the second derivative of the function f(x) given by f(x)=e^2x+sin2x/2x ?

To determine the second derivative, we'll have to
determine the 1st derivative, for the beginning. We'll differentiate f(x) with respect
to x.

f'(x) =
(e^2x+sin2x/2x)'

We'll apply chain rule for the first term
of the sum and the quotient rule for the 2nd terms of the
sum.

f'(x) = 2e^2x + [(sin 2x)'*2x - (sin
2x)*(2x)']/4x^2

f'(x) = 2e^2x + [2*(cos 2x)*2x - 2*(sin
2x)]/4x^2

f'(x) = 2e^2x + 2*(2x*cos 2x - sin
2x)/4x^2

f'(x) = 2e^2x + (2x*cos 2x - sin
2x)/2x^2

Now, we'll determine the second
derivative.

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

f"(x) = [2e^2x +
(2x*cos 2x - sin 2x)/2x^2]'

f"(x) = 4e^2x + [(2x*cos
2x)/2x^2]' - [(sin 2x)/2x^2]'

f"(x) = 4e^2x + (cos 2x/x)' -
(4x^2*cos 2x - 4x*sin 2x)/4x^4

f"(x) = 4e^2x + (-2xsin 2x -
cos 2x)/x^2 - 4x^2*cos 2x/4x^4 + 4x*sin 2x/4x^4

f"(x) =
4e^2x - 2*sin 2x/x - cos 2x/x^2 - cos 2x/x^2 + sin
2x/x^3

f"(x) = 4e^2x - 2*sin 2x/x - 2*cos 2x/x^2 + sin
2x/x^3

The second derivative is: f"(x) =
4e^2x - 2*sin 2x/x - 2*cos 2x/x^2 + sin 2x/x^3.