How do you differentiate #f(x) = (sinx)/(sinx-cosx)# using the quotient rule?

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Answer 1 · 2016-01-13T15:18:35

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)#