Another way of writing this:
(x2+2x+1+3x-1)/(x2+2x+1).
So we get 1+[(3x-1)/(x2+2x+1)], which is the same as 1+[(3x-1)/(x+1)2].
Since the derivative of the constant 1 is zero we need only differentiate (3x-1)/(x+1)2:
(3(x+1)2-2(3x-1)(x+1))/(x+1)4, which can be written:
(3x+3-6x+2)/(x+1)3=(5-3x)/(x+1)3, because the factor x+1 is common to numerator and denominator.
The derivative can be used to find critical points on the graphed curve. The first point to note is that x=-1 is an asymptote, and when x is close to -1, x(5+x)/(x2+2x+1) becomes large and negative. When the derivative is zero, 5=3x, so x=5/3 is a turning point (5/3,25/16). Note that 25/16>1. When x=0 the graph passes through the origin from being negative so the behaviour of the curve is that it rises to a maximum when x=5/3. This is a critical point which helps to sketch the graph. At x=0 the derivative=5, which is the steep positive gradient of the curve as it passes through the origin. Note that there is no other turning point.
We also know that when x is large and positive or negative we can use the rewritten form of the graph: y=f(x)=1+[(3x-1)/(x+1)2] which gets closer to 1, because the term involving x gets smaller and smaller, so there is a horizontal asymptote y=1. This asymptote applies to both parts of the graph on either side of x=-1. Because the maximum was at (5/3,25/16), that is, higher than the asymptote at x=1, it indicates that the curve falls towards the asymptote, while the other part of the graph (x<-1) rises towards it as x gets more and more negative. There is no turning point here.
The derivative is part of the graphing strategy enabling you to sketch the curve.