>>11048913You need to use both, often times using only one tool won't be enough to cut it, especially once you start seeing uglier functions. If you have a product with two expressions that contain the variable you are trying to differentiate with respect to (so x), it should be obvious by now that you need to use the product rule (or quotient, depending on how you wanna see it). Here you have a product of x and (x+1)^-1, hence you start with product rule:
(x)'[(x+1)^-1]+(x)[(x+1)^-1]'
To compute [(x+1)^-1]' you'll need the chain rule, although once you see it enough times you'll probably do it automatically. The gist of the chain rule is that it is used when you have a function "inside" of another function (which is NOT the same as a product or another standard algebraic operation). If you consider the functions f(x)=(x+1) and g(x)=x^-1, you'll see that (x+1)^-1 can be written as g(f(x)) (so f "inside" x, f(x) is what gets evaluated in g). Then the chain rule says that the derivative is g'(f(x))*f'(x), or:
[(x+1)^-1]'=[(-1)((x+1)^-2)]*(1)= -(x+1)^-2
And then you finish the sum, since you know both the derivatives you need
(x)'[(x+1)^-1]+(x)[(x+1)^-1]'
[(x+1)^-1]-x(x+1)^-2, after rewriting it around a little
=1/(x+1)^2
