The Quotient Rule States
#[(f(x))/(g(x))]'=(f'(x)g(x)-f(x)g'(x))/(g(x)^2)#
In this case
#f(x)=sqrt(x+1)#
and
#g(x)=x^2#
To find #f'(x)# we will use the Chain Rule, which states
#[h(k(x))]'=h'(k(x))k'(x)#
In this case
#h(x)=sqrt(x)#
and
#k(x)=x+1#
Since the derivative of a constant is #0# and the derivative of #x# is #1#
#k'(x)=1#
Also remember that #sqrt(x)=x^(1/2)#
and the Power rule tells us that #(x^n)'=nx^(n-1)#
Then
#h'(x)=1/2x^(1/2-1)=1/2x^(-1/2)=1/(2sqrt(x)#
Then we plug in and find that
#f'(x)=1/(2sqrt(x+1))#
and we can use the power rule again to find #g'(x)#
#g'(x)=(x^2)'=2x^(2-1)=2x^1=2x#
Then we plug in
#(sqrt(x+1)/(x^2))'=((x^2)/(2sqrt(x+1))-2xsqrt(x+1))/(x^2)^2#
and we simplify
#=((x^2)/(2sqrt(x+1))-2xsqrt(x+1))/(x^4)#
#=((x^2)/(2sqrt(x+1)))x^-4-(2xsqrt(x+1))/x^4#
#=x^-2/(2sqrt(x+1))-(2sqrt(x+1))/x^3#
#=1/(2x^2sqrt(x+1))-(2sqrt(x+1))/x^3#
