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Differentiate the function with respect to x: (log x)log x, x > 1

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प्रश्न

Differentiate the function with respect to x:

(log x)log x, x > 1

योग
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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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अध्याय 5: Continuity and Differentiability - Exercise 5.9 [पृष्ठ १९१]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 5 Continuity and Differentiability
Exercise 5.9 | Q 7 | पृष्ठ १९१

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