How do you find the derivative of #((sinx)^2)/(1-cosx)#?

1 Answer
Jun 17, 2016

Answer:

#-sinx#

Explanation:

The derivative of the quotient #u/v#

#d(u/v)=(u'v-v'u)/v^2#

Let #u=(sinx)^2# and #v=1-cosx#

#(d(sinx)^2)/dx=2sin(x)*(dsinx)/dx#

#=2sinxcosx#

#color(red)(u'=2sinxcosx)#

#(d(1-cos(x)))/dx=0-(-sinx)=sinx#

#color(red)(v'=sinx)#

Apply the derivative property on the given quotient:

#(d(((sinx)^2)/(1-cosx)))/dx#

#=((2sinxcosx)(1-cosx)-sinx(sinx)^2)/(1-cosx)^2#
#=((2sinxcosx)(1-cosx)-sinx(1-(cosx)^2))/(1-cosx)^2#
#=((2sinxcosx)(1-cosx)-sinx(1-cosx)(1+cosx))/(1-cosx)^2#
#((1-cosx)[2sinxcosx-sinx(1+cosx)])/(1-cosx)^2#

Simplify by #1-cosx# this leads to

#=(2sinxcosx-sinx(1+cosx))/(1-cosx)#
#=(2sinxcosx-sinx-sinxcosx)/(1-cosx)#
#=(sin xcosx-sinx)/(1-cosx)#
#=(-sinx(-cosx+1))/(1-cosx)#
#=(-sinx(1-cosx))/(1-cosx)#
Simplify by #1-cosx#

#=-sinx#