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

### 2 Answers | Add Yours

We have the function f(x) = e^2x + (sin 2x) / 2x

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

=> (e^2x)' + [(sin 2x/2x]'

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

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

f''(x) = (f'(x))'

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

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

=> 4*e^2x - (cos 2x + 2*x*sin 2x)/x^2 - (1/2)*[(sin 2x)*2x + 2*x^2* cos 2x)/x^4]

=> 4*e^2x + (cos 2x + 2*x*sin 2x)/x^2 - [(sin 2x + x*cos 2x)/x^3]

**The required derivative is 4*e^2x + (cos 2x + 2*x*sin 2x)/x^2 - (sin 2x + x*cos 2x)/x^3**

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.**