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

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We have to find the second derivative of y = e^5x + (ln x) / 2x

y = e^5x + (ln x) / 2x

=> y = e^5x + (1/2)(ln x)*(1/x)

y' = 5*e^5x + (1/2)[(-1)(x^-2)(ln x) + (1/x^2)]

y'' = 25*e^5x + (1/2)[2*(x^-3)*ln x - x^-3 - 2/x^3]

y'' = 25*e^5x + (1/2)[2*ln x / x^3 - 3/ x^3]

y'' = 25*e^5x + (ln x) / (x^3) - (3/2)*(1/x^3)

**The required second derivative is: 25*e^5x + (ln x)/(x^3) - (3/2)*(1/x^3)**

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.**