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प्रश्न
If y = `(sin x)^sin x` , then `"dy"/"dx"` = ?
विकल्प
`(sin x)^(cos x) {sin log [sin (pi/4 - x/2)]}`
`(sin x)^(cos x) {sin log [cos (pi/4 + x/2)]}`
(sin x)sin x [cos x. log (sin x) - cos x]
(sin x)sin x [cos x. log (sin x) + cos x]
MCQ
व्युत्पत्ति
उत्तर
(sin x)sin x [cos x. log (sin x) + cos x]
Explanation:
y = `(sin x)^sin x`
Taking logarithm on both sides, we get
log y = sin x log (sin x)
Differentiating both sides w.r.t. x, we get
`1/y * "dy"/"dx" = sin x * 1/(sin x) * cos x + log (sin x) * cos x`
`=> 1/y * "dy"/"dx" = cos x + cos x log (sin x)`
`=> "dy"/"dx" = (sin x)^(sin x) [cos x + cos x log (sin x)]`
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