The idea is to find the slope of the curve at x by adding a little bit to x, (x + h), calculating f(x + h), then subtract f(x), and finally divide by h and finding the limit as h gets very small. It works great for y = x

^{2}. But consider

y = x |

We want to do:

1/h [(x + h) |

So thinking about the binomial expansion of the x + h term, I'm trying to figure what

*is*the expansion for the 1/2 power? I mean, what can "n choose k" mean when n is 1/2?

The other thing that is a good approach here is to bring x out of the sum as follows

(x + h) |

To expand this, we just remember Newton's super-duper binomial expansion from here. If the expression is

(1 + Q) |

the expansion is

1 + (m/n)Q + (m/n)((m-n)/2n)Q |

We only need the first two terms because Q

^{2}is (h/x)

^{2}which we will ignore. So we have for the expansion

(1 + h/x) |

and the derivative is the limit as h => 0 of

1/h [x |

Similarly for m = -2 and n = 1 the first two terms are 1 and -2 and the expansion we need will be

(1 + (h/x)) |

The derivative is the limit as h => 0 of

1/h [x |

Easy when you know how.

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