A Look at Why Anti-Derivatives May be Tougher than Derivatives

 The definition of the derivative is the limit as x goes to zero of (F(x+dx)-F(x))/dx.  If you're given a function, it's not too bad to plug it into this thing, do some algebra, take the limit, and see what happens.

But try going backwards.  Try taking a function, then doing algebraic manipulations to get another function into this form.  Seems to be much tougher, and I guess there are a lot more potential options you could do, but then getting them to simplify in the right way seems somewhat daunting if you're like me and not all that great at algebraic manipulation.

Let's take something simple like 2x for example, and try getting its anti-derivative by algebraically manipulating it to look like something that fits the definition of the derivative.

2x

First let's add dx to it.  Since dx will just be going to zero when the limit is taken, I guess it's kind of like adding zero.

2x+dx

Then multiply by 1, but put 1 in the form of dx over dx.

(2x+dx)(dx/dx)

Multiply the top dx through.

(2xdx+dx^2)/dx

Add zero to the numerator, with zero in the form of x squared minus itself.

(2xdx+dx^2+x^2 – x^2)/dx

Shift one of the x squareds to the front.

(x^2+2xdx+dx^2 – x^2)/dx

Realize the first three terms can combine.

((x+dx)^2 – x^2)/dx

Then realize that we now have the function x^2 in the form of the limit as x goes to zero of (F(x+dx)-F(x))/dx.

This implies that the derivative of x^2 is 2x, which means the anti-derivative of 2x is x^2.

That was a LOT of algebraic manipulation I would have NEVER thought of doing if I didn't do this problem backwards to begin with.  And this is one of the easiest anti-derivatives.  Imagine if we tried something harder.