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- anonymousLv 71 month agoFavourite answer
y = ln[ln(x)]

Finding the first derivative involves the differentiation chain rule.

y' = [1 / ln(x)] * (d/dx) ln(x)

= [1 / ln(x)] * (1/x)

= 1 / x ln(x)

Finding the second derivative involves the differentiation quotient and product rules.

y'' = { [x ln(x) * (d/dx) 1] - [1 * (d/dx) x ln(x)] } / [x ln(x)]²

= { [x ln(x) * 0] - [1 * (d/dx) x ln(x)] } / x² ln²(x)

= - (d/dx) x ln(x) / x² ln²(x)

= - { [x * (d/dx) ln(x)] + [ln(x) * (d/dx) x] } / x² ln²(x)

= - { [x * (1/x)] + [ln(x) * 1] } / x² ln²(x)

= - [1 + ln(x)] / x² ln²(x)

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