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प्रश्न
(sin x)cosx
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उत्तर
Let y = (sin x)cosx
⇒ log y = log(sin x)cosx = cosx log(sinx)
Differentiate both sides w.r.t. x, we get
⇒ `1/y * "dy"/"dx" = cos x * "d"/"dx" (log sin x) + log sin x "d"/"dx" (cos x)`
= `cos x * 1/sinx * "d"/"dx" (sin x) + log sin x * (- sin x)`
= `cot x * cos x - log (sin x) * sin x`
∴ `"dy"/"dx" = y[cot x cos x - sin x * log(sin x)]`
= (sin x)cosx [cot x cos x – sin x · log(sin x)]
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