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Differentiate the function with respect to x. (log x)^cos x

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प्रश्न

Differentiate the function with respect to x.

(log x)cos x

योग
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उत्तर

Let, y = (log x)cos x

Taking logarithm of both sides,

log y = log (log x)cos x

= cos x log (log x)  ...[∵ log mn = n log m]

Differentiating both sides with respect to x,

`1/y dy/dx = cos x d/dx log (log x) + log (log x) d/dx cos x`

`1/y dy/dx = cos x * 1/(log x) d/dx (log x) + log (log x) (- sin x)`

`1/y dy/dx = cos x * 1/(log x) * 1/x - sin x log (log x)`

`1/y dy/dx = - sin x log (log x) + (cos x)/(x log x)`

`dy/dx = y [- sin x log (log x) + (cos x)/(x log x)]`

`dy/dx = (log x)^(cos x) [- sin x log (log x) + (cos x)/(x log x)]`

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अध्याय 5: Continuity and Differentiability - Exercise 5.5 [पृष्ठ १७८]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 5 Continuity and Differentiability
Exercise 5.5 | Q 3 | पृष्ठ १७८

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