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

Question

1474 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-30T04:18:57

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

For a function #f(x)=(g(x))/(h(x))#,

#f'(x)=(g'(x)h(x)-g(x)h'(x))/(h(x))^2#

Here,

#g(x)=cosx#
#h(x)=sinx-cosx#

and

#g'(x)=-sinx#
#h'(x)=cosx+sinx#

so,

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

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

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