How do you find the derivative of# sinx/(1+cosx)#?

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Answer 1 · 2018-05-03T17:24:19

#1/(cosx+1)#

#f(x)=sinx/(cosx+1)#

#f'(x)=(sinx/(cosx+1))'#

The derivative of #f(x)/g(x)# using Quotient Rule is
#(f'(x)g(x)-f(x)g'(x))/g^2(x)#
so in our case it is

#f'(x)=((sinx)'(cosx+1)-sinx(cosx+1)')/(cosx+1)^2# #=#

#(cosx(cosx+1)+sin^2x)/(cosx+1)^2# #=#

#(color(blue)(cos^2x)+cosx+color(blue)(sin^2x))/(cosx+1)^2# #=#

#cancel((cosx+color(blue)(1)))/(cosx+1)^cancel(2)# #=#

#1/(cosx+1)#

Answer 2 · 2018-05-03T18:08:38

#1/2sec^2(x/2) or 1/(1+cosx)#.

We have, #sinx/(1+cosx)#,

#={2sin(x/2)cos(x/2)}/{2cos^2(x/2)}#,

#=tan(x/2)#.

#"Therefore, "d/dx{sinx/(1+cosx)}#,

#=d/dx{tan(x/2)}#,

#=sec^2(x/2)*d/dx{x/2}......"[The Chain Rule]"#,

#=sec^2(x/2)*1/2#,

#=1/2sec^2(x/2), or, #

#=1/(2cos^2(x/2))#,

#=1/(1+cosx)#.