d/dx of sinx sin2x sin3x give me the answer please derivation with step by step

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Saurabh Koranglekar · Answer

Dear student Please find the link below https://www.askiitians.com/forums/Differential-Calculus/find-dy-dx-when-y-sin-x-sin-2x-sin-3x-sin-4x_101593.htm Regards

Aditya Gupta · Answer

y= sinxsin2xsin3x lny= lnsinx + lnsin2x + lnsin3x Differentiate both sides wrt x using chain rule y'/y= cosx/sinx + 2cos2x/sin2x + 3cos3x/sin3x Hence, y'= cosxsin2xsin3x + 2cos2xsinxsin3x + 3cos3xsinxsin2x. KINDLY APPROVE :))

Arun · Answer

Dear student For differentiation, we should use chain rule dy/dx = cos x sin2x sin3x + 2 sinx cos2x sin3x + 3 sinx sin2x cos3x Hope it is helpful

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d/dx of sinx sin2x sin3x give me the answer please derivation with step by step

d/dx of sinx sin2x sin3x give me the answer please derivation with step by step

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