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प्रश्न
`2^(cos^(2_x)`
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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