How do you differentiate #f(x)=sinx/(1-cosx)#?

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Answer 1 · 2016-10-05T10:03:23

#(df)/(dx)=-1/(1-cosx)#

We use the quotient formula here. It states if #f(x)=(g(x))/(h(x))#

then #(df)/(dx)=((dg)/(dx)xxh(x)-(dh)/(dx)xxg(x))/(h(x))^2#

As #f(x)=sinx/(1-cosx)#

#(df)/(dx)=(cosx xx(1-cosx)-sinx xxsinx)/(1-cosx)^2#

= #(cosx-cos^2x-sin^2x)/(1-cosx)^2#

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

= #(cosx-1)/(1-cosx)^2#

= #-(1-cosx)/(1-cosx)^2#

= #-1/(1-cosx)#