Question

How did they get `6csc2x`?

The question was find the derivative of `y=(sin^2xtan^3x)/(x^2+1)^2`. I got `(2cotx+(3sec^2)/tanx-(4x)/(x^2+1))(sin^2xtan^3x)/(x^2+1)^2` as my solution. The correct answer is `(2cotx+6csc2x-(4x)/(x^2+1))(sin^2xtan^3x)/(x^2+1)^2`

Answer 1

`y'=(2cotx+(3sec^2x)/tanx-(4x)/(x^2+1))(sin^2xtan^3x)/(x^2+1)^2` and `y'=(2cotx+6csc2x-(4x)/(x^2+1))(sin^2xtan^3x)/(x^2+1)^2` are same and correct.

Explanation:

Your Question is
`y=(sin^2xtan^3x)/(x^2+1)^2`

and your answer is
`y'=(2cotx+(3sec^2x)/tanx-(4x)/(x^2+1))(sin^2xtan^3x)/(x^2+1)^2`
there is a little bit of transformation
write `sec^2x` as `1/cos^2x` and `tanx` as `sinx/cosx`
`y'=(2cotx+(3(cosx))/(cos^2xsinx)-(4x)/(x^2+1))(sin^2xtan^3x)/(x^2+1)^2`
`y'=(2cotx+(3)/(cosxsinx)-(4x)/(x^2+1))(sin^2xtan^3x)/(x^2+1)^2`
`y'=(2cotx+(3)(2)/(2cosxsinx)-(4x)/(x^2+1))(sin^2xtan^3x)/(x^2+1)^2`
use Formula `sin2x=2sinxcosx`
we get
`y'=(2cotx+(6)/(sin2x)-(4x)/(x^2+1))(sin^2xtan^3x)/(x^2+1)^2`
write `sin2x` as `1/csc(2x)`
`y'=(2cotx+6csc2x-(4x)/(x^2+1))(sin^2xtan^3x)/(x^2+1)^2`

and you get the answer.