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''...

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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)

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