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Question
Find the second order derivatives of the following : log(logx)
Solution
Let y = log(logx)
Then `"dy"/"dx" = "d"/"dx"[log (logx)]`
= `(1)/"logx" . "d"/"dx"(logx)`
= `(1)/"logx" xx (1)/x = (1)/"xlogx"`
and
`(d^2y)/(dx^2) = "d"/"dx"(xlogx)^-1`
= `(-1)(xlogx)^-2."d"/"dx"(xlogx)`
= `(-1)/(xlogx)^2.[x"d"/"dx"(logx) + (logx)."d"/"dx"(x)]`
= `(-1)/(xlogx)^2.[x xx 1/x + (logx) xx 1]`
= `-(1 + logx)/(xlogx)^2`.
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