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

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

$y=u/v$

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

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

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

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

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

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

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