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Q.Find the derivatives (any two): sin⁡xsin⁡2x\sin x \sin 2x

Assam AhsecAHSEC Higher Secondary (HS) 1st Year Examination 2018Subjective· 2mImportance★★★★★
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By the product rule, the derivative is cos⁡xsin⁡2x+2sin⁡xcos⁡2x\cos x\sin2x+2\sin x\cos2x.

By the product rule with u=sin⁡xu=\sin x, v=sin⁡2xv=\sin2x:

ddx(sin⁡xsin⁡2x)=(cos⁡x)(sin⁡2x)+(sin⁡x)ddx(sin⁡2x).\dfrac{d}{dx}(\sin x\sin2x)=(\cos x)(\sin2x)+(\sin x)\dfrac{d}{dx}(\sin2x).

By the chain rule, ddx(sin⁡2x)=2cos⁡2x\dfrac{d}{dx}(\sin2x)=2\cos2x. So:

ddx(sin⁡xsin⁡2x)=cos⁡xsin⁡2x+2sin⁡xcos⁡2x.\dfrac{d}{dx}(\sin x\sin2x)=\cos x\sin2x+2\sin x\cos2x. …

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