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Answer 1 · 2018-01-19T02:58:14

$y'=1/(x+1)^2$

$y'=d/dx(f(x)/g(x))=(d/dx[f(x)]g(x)-d/dx[g(x)]f(x))/(g(x))^2$

Given: $(2x+1)/(x+1)$

Let

$f(x)=2x+1=>d/dx[f(x)]=2$

$g(x)=x+1=>d/dx[g(x)]=1$

Substituting into the quotient rule

$y'=((2)(x+1)-(1)(2x+1))/(x+1)^2$

Simplifying:

$y'=((2x+2)-(2x+1))/(x+1)^2=((cancel(2x)+2)-(cancel(2x)+1))/(x+1)^2=1/(x+1)^2$