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  • 3 years ago

OK, here is a differentiation problem that is baffling me -- y = {ln[ln(ln x)]} -- What is y prime? I am guessing the chain rule is the key here, but I cannot figure out the solution.

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  1. marsss
    • 3 years ago
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    [ln(ln x)] = z and ln x = t => z = ln t y = ln z y(prime) = z(prime)/z z(prime) = t(prime)/t t(prime) = 1/x y(prime) = 1/xtz y(prime) = 1/x*ln x*ln(ln x)

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