How do you differentiate #f(x) = (sinx)/(sinx-cosx)# using the quotient rule?
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The answer is:
#f'(x)=-cosx(sinx+cosx)/(1-sin2x)#
The quotent rule states that:
#a(x)=(b(x))/(c(x))#
Then:
#a'(x)=(b'(x)*c(x)-b(x)*c'(x))/(c(x))^2#
Likewise for #f(x)#:
#f(x)=(sinx)/(sinx-cosx)#
#f'(x)=((sinx)'(sinx-cosx)-sinx(sinx-cosx)')/(sinx-cosx)^2#
#f'(x)=(cosx(sinx-cosx)-sinx(cosx-(-cosx)))/(sinx-cosx)^2#
#f'(x)=(cosxsinx-cos^2x-sinxcosx-sinxcosx)/(sinx-cosx)^2#
#f'(x)=(-sinxcosx-cos^2x)/(sinx-cosx)^2#
#f'(x)=-cosx(sinx+cosx)/(sinx-cosx)^2#
#f'(x)=-cosx(sinx+cosx)/(sin^2x-2sinxcosx+cos^2x)#
#f'(x)=-cosx(sinx+cosx)/((sin^2x+cos^2x)-2sinxcosx)#
#f'(x)=-cosx(sinx+cosx)/(1-sin2x)#
