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Question
Find the second order derivative of the function.
log (log x)
Solution
Let, y = log (log x)
Differentiating both sides with respect to x,
`dy/dx = d/dx log (log x) = 1/(log x). d/dx (log x) = 1/(log x) xx 1/x`
`therefore dy/dx = 1/(x log x) = (x log x)^-1`
`(d^2y)/dx^2 = 1/logx (-1/x^2) + 1/x d/dx (1/log x)`
`= (-1)/(x^2 log x) + 1/x [(logx. 0 - 1 . 1/x)/(log x)^2]`
`= (-1)/ (x^2 log x) + 1/x [(-1/x)/(logx)^2]`
`= (-1)/ (x^2 log x) - 1/ (x^2(logx)^2)`
`= (-1)/ (x^2 log x) [1 + 1/ logx]`
`= (-1 (1 + log x))/ (x log x)^2`
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