Originally posted by joe shmo
That makes my head hurt...
And I don't speak the language of combinatorics. I'll try to learn it soon..
until then, care to give me something less analytical...ie more verbal(dumbed down) 🙂
Sure, I'll give it a shot. Starting with the function f(x) = x^n, we find the derivative by applying the definition:
f'(x) = lim (h-->0) (f(x+h) - f(x)) / h = lim (h-->0) ((x+h)^n - x^n) / h
Let's start with a few simple examples, and determine the pattern from there. Starting with n=2 we have:
f'(x) = lim (h-->0) ((x+h)^2 - x^2) / h
Dropping the limit notation temporarily (for clarity), we continue with the algebra:
= (x^2 + 2xh + h^2 - x^2) / h
= (2xh + h^2) / h
= h * (2x + h) / h
= 2x + h
Now, lim (h-->0) 2x + h = 2x. Therefore, f'(x) = (x^2)' = 2x. So far so good! Let's try it with n=3:
f'(x) = lim (h-->0) ((x+h)^3 - x^3) / h
Again, dropping the limit notation temporarily, we have:
= (x^3 + 3hx^2 + 3xh^2 + h^3 - x^3) / h
= (3hx^2 + 3xh^2 + h^3) / h
= h * (3x^2 + 3xh + h^2) / h
= 3x^2 + 3xh + h^2
Now, lim (h-->0) 3x^2 + 3xh + h^2 = 3x^2. A pattern is emerging! Let's try one more time, with n=4:
f'(x) = lim (h-->0) ((x+h)^4 - x^4) / h
Again, dropping the limit notation temporarily, we have:
= (x^4 + 4hx^3 + 6(h^2)(x^2) + 4xh^3 + h^4 - x^4) / h
= (4hx^3 + 6(h^2)(x^2) + 4xh^3 + h^4) / h
= h * (4x^3 + 6hx^2 + 4xh^2 + h^3) / h
= 4x^3 + 6hx^2 + 4xh^2 + h^3
Now, lim (h-->0) 4x^3 + 6hx^2 + 4xh^2 + h^3 = 4x^3. Let's review:
(x^2)' = 2x
(x^3)' = 3x^2
(x^4)' = 4x^3
This pattern suggests that for any n, (x^n)' = nx^(n-1), which is easily explained by the combinatorial nature of the binomial expansion. What's that you say? Let me explain.
Say we start with (x+h)^n. This is just short form notation for (x+h)(x+h)(x+h)...(x+h), where there are "n" (x+h)'s in a row. In order to expand this expression, we multiply it out and group the results by variable type (i.e. x^2, hx^4, (h^3)(x^2), etc...). However, a nifty way to think about generating each term is to imagine "choosing" either an "x" or an "h" from each bracket. This "choosing" process will generate each variable type in order, and it will generate the coefficients as well. Handy! For instance, say we want to generate the term hx^(n-1). How many ways can you choose one h and the rest x's? Well, since there are "n" brackets to choose from, there will be "n" ways to choose that h. The rest must be x's, as stated. Therefore, the term will have the form n*hx^(n-1). What do we find in the examples above? When n=2, the coefficient for xh is 2. When n=3, the coefficienct for hx^2 is 3. When n=4, the coefficient for hx^3 is 4. Great! (Also note that the sum of the exponents in each case always adds up to "n", because for every term we must make a choice of either x or h for each of the "n" brackets.) We can repeat this choosing process for each term, resulting in a short form expression using choose notation which is used in the Wikipedia article I posted earlier.
For reference, CHOOSE(n,k) = n!/((n-k)!*k!).
So how does all this apply to the general power rule? Well, from the definition of the derivative, we know that the x^n term from the expansion of the polynomial will cancel with the x^n term from the definition. The remaining terms must all have at least one "h". We can factor out this "h", which will cancel with the "h" in the denominator, leaving us with nx^(n-1) and several more terms all still containing at least one "h". Since the limit of any term containing an "h" is 0 as h-->0, we are simply left with nx^(n-1). Voila!
The Wikipedia article I posted fleshes this out with more rigour, using short form notation that can get complicated quickly. I hope this explanation is a little easier to absorb. 🙂
20 Aug '09 18:11
calc question
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