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

Sravan kumar , 5 Years ago
Grade 11
anser 3 Answers
Saurabh Koranglekar

Last Activity: 5 Years ago

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

Last Activity: 5 Years ago

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

Last Activity: 5 Years ago

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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