How do you differentiate #f(x)= x/(4x-2)# using the quotient rule?

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Answer 1 · 2016-01-26T13:02:40

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

#f(x)=x/(4x-2)=1/2(x/(2x-1))#

We could taking 2 at the denominator to make easy the computation

#f'(x)=d/(dx)[1/2(x/(2x-1))]=1/2d/(dx)(x/(2x-1))=1/2d/(dx)i(x)#

Now #i(x)=(h(x))/g(x)#

The Quotient Rule tells us:

given: #i(x)=(h(x))/g(x)#

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

#:.f'(x)=1/2[(1*(2x-1)-(2-0)*x)/(2x-1)^2]=#

#=1/2[(color(green)cancel(2x)-1color(green)cancel(-2x))/(2x-1)^2]=#

#=1/2*(-1/(2x-1)^2)=-1/(2(2x-1)^2#