What is the second derivative of y=e^5x+(lnx)/2x ?

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

dy/dx =
d(e^5x+(lnx)/2x)/dx

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

dy/dx = 5e^5x + [(lnx)'*2x -
(lnx)*(2x)']/4x^2

dy/dx = 5e^5x + (2x/x -
2lnx)/4x^2

dy/dx = 5e^5x + 2(1 -
lnx)/4x^2

dy/dx = 5e^5x + (1 -
lnx)/2x^2

dy/dx = 5e^5x + (lne -
lnx)/2x^2

dy/dx = 5e^5x +
[ln(e/x)]/2x^2

Now, we'll determine the second
derivative:

d^2y/dx^2 = [d[5e^5x +
[ln(e/x)]/2x^2/dx

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

d^2y/dx^2= 25e^5x + {-2x^2/x -
[1-ln(x]*4x}/4x^4

d^2y/dx^2 = 25e^5x -
2x(1+2-2lnx)/4x^4

d^2y/dx^2 = 25e^5x -
2x(1+2-2lnx)/4x^4

d^2y/dx^2 = 25e^5x -
(1+2-2lnx)/2x^3

d^2y/dx^2 = 25e^5x - (3 -
lnx^2)/2x^3

The second derivative is:
d^2y/dx^2 = 25e^5x - (3 - lnx^2)/2x^3.