Question

How do you find the second derivative of `ln^2 (x)`?

Answer 1

`:. (d^2y)/(dx^2)={2(1-lnx)}/x^2=ln{(e/x)^(2/x^2)}.`

Explanation:

Let `y=ln^2(x)=(lnx)^2`.

`:. dy/dx=2lnx*d/dx(lnx)`.......[Chain Rule].....`=(2lnx)/x.`

`:. (d^2y)/(dx^2)=d/dx{dy/dx}`........................[Defn.]
`=d/dx{(2lnx)/x}`
`=2[{x*d/dx(lnx)-(lnx)*d/dx(x)}/x^2]`..........[Quotient Rule]
`=2[{x*1/x-(lnx)(1)}/x^2]`
`={2(1-lnx)}/x^2`
`=(2/x^2)(lne-lnx)`
`=(2/x^2)*ln(e/x)`
`=ln{(e/x)^(2/x^2)}`

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