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d/dx of sinx sin2x sin3x give me the answer please derivation with step by step
Dear studentPlease find the link belowhttps://www.askiitians.com/forums/Differential-Calculus/find-dy-dx-when-y-sin-x-sin-2x-sin-3x-sin-4x_101593.htmRegards
y= sinxsin2xsin3xlny= lnsinx + lnsin2x + lnsin3x Differentiate both sides wrt x using chain ruley'/y= cosx/sinx + 2cos2x/sin2x + 3cos3x/sin3xHence,y'= cosxsin2xsin3x + 2cos2xsinxsin3x + 3cos3xsinxsin2x.KINDLY APPROVE :))
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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