Question

find nth derivative of sinx cos2x sin3x​

Answer 1

we can differentiate by parts twice. Let u=sinx and v=sin2xsin3x. Differentiate uv and we have udv/dx+vdu/dx, that is: sinxd(sin2xsin3x)/dx+sin2xsin3xcosx. Now we differentiate by parts again: let p=sin2x and q=sin3x, so differentiating pq we have: sin2x.3cos3x+sin3x.2cos2x=3sin2xcos3x+2cos2xsin3x. Put this in place of d(sin2xsin3x)/dx:

sin(3sin2xcos3x+2cos2xsin3x)+sin2xsin3xcosx=3sinxsin2xcos3x+2sinxcos2xsin3x+cosxsin2xsin3x.

That completes the differentiation, but the trig functions could be combined, perhaps because

sin(A+B)=sinAcosB+cosAsinB, and

sin(A+B+C)=sinA(cosBcosC-sinBsinC)+cosA(sinBcosC+cosBsinC),

and if A=B=C=x then we can calculate sin3x and sin2x. It's arguable whether such substitutions would make the derivative more complicated or less complicated! If I have time, perhaps I could try it, or you could try it? But I think the question wants you to focus on the derivative, using the product rule, rather than play around with trig functions.

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$\huge\red{\ddot\smile}$ hope it helps you ✅✔️

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