If #f(x) = (3x-2)/(2x+1)# what is #f'(x)#?

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Answer 1 · 2017-03-11T12:27:43

The answer is #=7/(2x+1)^2#

This is the derivative of a quotient

#(u/v)'=(u'v-uv')/v^2#

Here,

#f(x)=(3x-2)/(2x+1)#

#u=3x-2#, #=>#, #u'=3#

#v=2x+1#, #=>#, #v'=2#

Therefore,

#f'(x)=(3(2x+1)-2(3x-2))/(2x+1)^2#

#=(6x+3-6x+4)/(2x+1)^2#

#=7/(2x+1)^2#