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Continuity And Differentiability

Question
CBSEENMA12034739

Differentiate the following w.r.t.x: square root of log space left parenthesis logx right parenthesis end root

Solution
Let space straight y equals square root of log space left parenthesis logx right parenthesis end root equals left square bracket log space left parenthesis logx right parenthesis right square bracket to the power of begin inline style 1 half end style end exponent
therefore dy over dx equals 1 half left square bracket log space left parenthesis logx right parenthesis right square bracket to the power of negative begin inline style 1 half end style end exponent straight d over dx left square bracket log space left parenthesis logx right parenthesis right square bracket equals fraction numerator 1 over denominator 2 square root of log space left parenthesis logx right parenthesis end root end fraction cross times fraction numerator 1 over denominator log space straight x end fraction cross times straight d over dx left parenthesis log space straight x right parenthesis
equals fraction numerator 1 over denominator 2 square root of log space left parenthesis logx right parenthesis end root end fraction cross times fraction numerator 1 over denominator log space straight x end fraction cross times 1 over straight x equals fraction numerator 1 over denominator 2 xlog space straight x square root of log space left parenthesis logx right parenthesis end root end fraction

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