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Understanding the power rule?

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Suppose you want to differentiate f(x) = axⁿ

By definition, f’(x) is given by:
> lim(h→0) [f(x+h) - f(x)]/h

> = lim(h→0) [a(x+h)ⁿ - axⁿ]/h

> lim(h→0) a[(x+h)ⁿ - xⁿ]/h

Now, from the binomial theorem, we know that (x+h)ⁿ = xⁿ + nxⁿ⁻¹h +…+ nxhⁿ⁻¹ + hⁿ. Substituting that in, we see the xⁿ cancels out with the -xⁿ, leaving:
> lim(h→0) a(nxⁿ⁻¹h +…+ nxhⁿ⁻¹ + hⁿ)/h

Note that every term in the numerator now has an h in it, so we can factor that out and it’ll cancel out the h in the denominator:
> = lim(h→0) a(nxⁿ⁻¹ +…+ nxhⁿ⁻² + hⁿ⁻¹)

Plug in zero for all of the h’s to solve the limit:
> limit = a(nxⁿ⁻¹ +…+ nx(0)ⁿ⁻² + (0)ⁿ⁻¹)

> = a(nxⁿ⁻¹ + 0 + … + 0 + 0)

> **f’(x) = anxⁿ⁻¹**

So the general formula of the power rule comes directly from the limit-definition of the derivative. Once you understand this, you can skip the limit and just use the power rule as a shortcut.
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>my biggest problem is, how does that have anything to do with the power rule????

That's what a derivative *is*. A derivative is the slope of a tangent line. If you want to take the derivative of x^(n), you are asking how to take the limit of the slope between two points on the curve y=x^(n) as the distance between them shrinks to zero.

Recall that, by definition, f'(x) = limit as h goes to 0 of [f(x+h)-f(x)]/h. That's what they're calculating, so what they get is the derivative.

If you don't want to go all the way back to the limit definition though, you can also justify the power rule using the *product* rule! Recall that

x^(n)=(x)(x)(x)...(x)(x) n times,

so by the product rule, we add up terms where one factor at a time is differentiated, and since (x)'=1,

[x^(n)]' = (1)(x)(x)...(x) + (x)(1)(x)...(x) + (x)(x)(1)...(x) + ... + (x)(x)(x)...(1)(x) + (x)(x)(x)...(x)(1), with n terms

= x^(n-1) + x^(n-1) + x^(n-1) + ... + x^(n-1) with n terms

= nx^(n-1).
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If you use the binomial theorem to expand (x + h)^n for natural number n, the x^n terms cancel out and once you divide by h the only term without an h is n\*x^(n-1). Since h approaches zero all other terms in the expansion become zero, thus d(x^(n))/dx = n\*x^(n-1). There’s a separate proof for any real number instead of natural number.
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Let's take the function y=x^2 as an example.  Suppose you want to find the slope (that is, the rise divided by the run or, to be mathy, m=(y2-y1)/(x2-x1)) at the point x=1.  Since y=x^2, y=1 so that we have the fixed point (1,1).  Now, we need a second point to compute rise/run. Choose, say, x=4, y=16. Since you now have two points, you can compute the slope. Unfortunately,  your line goes through two points. This is called a secant line, and is not a tangent line through the single point (1,1).  Now, let's try it again with x=3, y=9. But, that's still a secant line, not a tangent line.  Let's try again with x=2, y=4. Again, secant line, not tangent.  But, notice that as we move from x=4 to x=3 to x=2, we are approaching x=1, our fixed point. The power rule for x^2 is 2x and gives you the slope at x=1 exactly and is the slope of the line tangent to x=1,y=1.  It's not the slope of the secant line since it now involves only a single point.
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>how does that have anything to do with the power rule????

because d/dx x^(n) = nx^(n-1) is differentiation, and differentiation is about slopes of tangent lines at a point, which are defined using a limit.

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