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प्रश्न
Differentiate the function with respect to x:
(log x)log x, x > 1
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उत्तर
Let y = (log x)log x
Taking log on both sides, we get
log y = log x log (log x) ....(1)
Differentiating (1) both sides with respect to x, we get,
`1/y dy/dx = log x* 1/log x * 1/x + log (log x) * 1/x`
= `1/x * [1 + log (log x)]`
`dy/dx = (log x)^(log x) * 1/x * [1 + log (log x)]`, x > 1
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