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

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Answer 1 · 2016-04-24T20:03:34

#d/(dx) ((x+2)/(x+3)) = 1/(x+3)^2#

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

#d/(dx) ((x+2)/(x+3)) = d/(dx) (1-(x+3)^(-1)) = (x+3)^(-2) = 1/(x+3)^2#

Answer 2 · 2016-04-24T20:11:30

#(d y)/(d x)=1/(x+3)^2#

#y=u/v#

#y'=(u'*v-v'*u)/(v^2)#

#y=(x+2)/(x+3)#

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

#(d y)/(d x)=((x+3)-(x+2))/(x+3)^2#

#(d y)/(d x)=(cancel(x)+3-cancel(x)-2)/(x+3)^2#

#(d y)/(d x)=1/(x+3)^2#