मराठी

2cos2x

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प्रश्न

`2^(cos^(2_x)`

बेरीज
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उत्तर

Let y = `2^(cos^(2_x)`

Taking log on both sides, we get

log y = `log 2^(cos^(2_x)`

⇒ log y = `cos^2x * log 2`

Differentiating both sides w.r.t. x

⇒ `1/y * "dy"/"dx" = log 2* "d"/"dx" cos^2x`

⇒ `1/y * "dy"/"dx" = log 2 [2 cos x * "d"/"dx" cos x]`

⇒ `1/y * "dy"/"dx" = log 2 [2 cos x(-sin x)]`

⇒ `1/y * "dy"/"dx" = log 2 (- sin 2x)`

`"dy"/"dx" = - y * log 2 sin 2x`

Hence, `"dy"/"dx" = -2^(cos^2x) (log 2 sin 2x)`

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पाठ 5: Continuity And Differentiability - Exercise [पृष्ठ १०९]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
पाठ 5 Continuity And Differentiability
Exercise | Q 25 | पृष्ठ १०९

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