Question

nth derivative ofcos^2x sin^3x ​

Answer 1

cos(2x) = sin(3x) 

so

cos(2x) = sin(x)cos(2x) + cos(x)sin(2x)

or

cos(2x)- sin(x)cos(2x) = cos(x)(2sin(x)cos(x))

cos(2x)[1 - sin(x)] = 2cos2(x)sin(x) = 2[1 - sin2(x)] sin(x)

Hence

cos(2x)[1 - sin(x)] = 2[1 - sin(x)][1 + sin(x)] sin(x)

and thus

cos(2x) = 2 [1 + sin(x)] sin(x)

Note: You need to check that sin(x) = 1 is not a solution to the original problem.

Finally

1 - 2 sin2(x) = 2 sin(x) + 2 sin2(x)

or

4 sin2(x) +2 sin(x) - 1 = 0

Solve for sin(x).

Answer 2

Answer:

refer the attachment in the top