Prooving derivate of x^n

1. Dec 1, 2005

Werg22

It is asked from me to proove that dy/dx x^ n = nx^n-1 without using the binominal theorem... any ideas?

2. Dec 1, 2005

BerkMath

There are a couple ways to do this. First, right x^n in terms of e and ln, the exponential and natural log function respectively. Then simply apply the chain rule. You can also arrive at the formula if you differentiate both sides of the expression: int(x^(n-1))=x^n/n, however the logic is somewhat circular.

3. Dec 1, 2005

HallsofIvy

How about using induction and the product rule?

(Since this has nothing to do with "differential equations", I am moving it.)

4. Dec 2, 2005

JasonRox

Integrating it would be a good approach.

Use the idea that (x^n-a^n) = (x-a)(x^n-1 + x^n-2*a ... + a^n-1)

Use the basic definition of a limit, and you get the left side as I showed above, then re-write it as the one on the right side, than you can cross (x-a) out if you use the appropriate definition (x->a). Then the right side is simply a geometric series, so sum that up and you get na^(n-1).

And you are done.

You have a proof for the definition (h->0), which uses the Binomial Theorem, and (x->a) like above.

5. Dec 2, 2005

HallsofIvy

Theorem: $\frac{d x^n}{dx}= nx^{n-1}$ for every positive integer n.

When n= 1, this says that $\frac{d x}{dx}= 1x^0= 1$ which is true.

Assume that, for some k, $\frac{dx^k}{dx}= kx^{k-1}$

$x^{k+1}= x(x^k)$ so, by the product rule,
$$\frac{dx^{k+1}}{dx}= \frac{dx}{dx}x^k+ x\frac{dx^k}{dx}[/itex] Then [tex]\frac{dx^{k+1}}{dx}= x^k+ x(kx^k)= (k+1)x^k$$

By induction, then, $\frac{dx^n}{dx}= nx^{n-1}$ for every positive integer n.

You can use the quotient rule to show it is true for all negative integers as well, the chain rule to show it is true for n any rational number, and, finally, logarithmic differentiation to show it is true for n any real number.

6. Dec 2, 2005

matt grime

just divide by x-a and the LHS in the limit is the derivative, do not just "cross it out"

it is not a geometric series (except in the trivial sense), it is simply a sum x^{n-1}+x^{n-1}+..+x^{n-1} with n terms in it that are all the same.

7. Dec 2, 2005

JasonRox

I guess my explanation is bad. I was too lazy to use LaTeX.

This was another proof given in a very popular text by Stewart.

8. Dec 3, 2005

Tide

How about this? We know that $x^n = x x^{n-1}$ and using the chain rule it follows that

$$D_n = x^{n-1} + x D_{n-1}$$

where $D_n$ stands for

$$\frac {d}{dx} x^n$$

Now just follow the recursion on $D_{n-1}$ all the way down to $D_0 = 0$ and you obtain your result with n identical terms of $x^{n-1}$. QED

9. Dec 6, 2005

HallsofIvy

That doesn't use the "chain rule", it uses the "product rule"- which I did earlier in this thread.