How do you use the chain rule to differentiate #y=(cosx/(1+sinx))^5#?

1 Answer
Pankaj Solanki
Sep 20, 2017

#-5(cos^4x)/(1+sinx)^5#

Explanation:

#y=(cosx/(1+sinx))^5#
#dy/dx=d/dx((cosx/(1+sinx))^5)#
#=5(cosx/(1+sinx))^4*(((1+sinx)*(d/dx(cosx))-cosxd/dx(1+sinx))/(1+sinx)^2)#

#=5(cosx/(1+sinx))^4*(((1+sinx)*(-sinx)-cosx(cosx))/(1+sinx)^2)#

#=5(cosx/(1+sinx))^4*((-sinx-sin^2x-cos^2x)/(1+sinx)^2)#

#=5(cosx/(1+sinx))^4*((-sinx-(sin^2x+cos^2x))/(1+sinx)^2)#

#=5(cosx/(1+sinx))^4*((-sinx-1)/(1+sinx)^2)#

#=5(cosx/(1+sinx))^4*(-1)/(1+sinx)#
#=-5(cos^4x)/(1+sinx)^5#

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