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4 January, 00:10

Derivative of square root of (lnx)

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  1. 4 January, 00:37
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    Use the chain rule

    y = √[ln (x) ]=> y' = [1/2√ (lnx) ]*[ln (x) ]' = [1/2√ (lnx) ] * 1/x = 1 / [2x (√lnx) ]
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