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#

1 Answer
Pankaj Solanki
Sep 16, 2017

#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.

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