How do you find the derivative of #(x-3)/(2x+1)#?

2 Answers
Jim G.
Oct 5, 2017

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

Explanation:

#"differentiate using the "color(blue)"quotient rule"#

#"given "f(x)=(g(x))/(h(x))" then"#

#f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larr" quotient rule"#

#g(x)=x-3rArrg'(x)=1#

#h(x)=2x+1rArrh'(x)=2#

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

#color(white)(rArrf'(x))=7/(2x+1)^2#

TreeReader
Oct 5, 2017

#=7/(4x^2-4x+1)# or #7/(2x+1)^2#

Explanation:

Quotient rule: #(u'v )-(uv')# all divided by #v^2#, assuming the top line is #u# and bottom is #v#

In this function, #u= x-3# and #v=2x+1#

#:. u' = 1 " and " v' = 2#

So plugging that in gets:

#(1*(2x+1)-(x-3)*2)/(2x+1)^2#

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

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

#=7/(4x^2-4x+1)#

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