How do you differentiate #f(x)=(2x^2-x+1)/(2x-1)# using the quotient rule?

1 Answer
Maurizio Giaffredo
Nov 2, 2015

#f^'(x)=[4x^2-4x-1]/[(2x-1)^2]#

Explanation:

Quotient rule states that if #h# and #k# are two differentiable functions on #]a,b[#, then #forall x in ]a,b[# such that #k(x) ne 0# the following equality holds:
#[(h(x))/(k(x))]^'=[h^'(x)k(x)-h(x)k^'(x)]/(k^2(x))#

In our case #h(x)=2x^2-x+1# and #k(x)=2x-1#. The two derivatives are #h^'(x)=4x-1# and #k^'(x)=2#.
#f^'(x)=[(4x-1)(2x-1)-(2x^2-x+1)2]/[(2x-1)^2]=[8x^2-6x+1-4x^2+2x-2]/[(2x-1)^2]=[4x^2-4x-1]/[(2x-1)^2]#

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