Proving the Derivative of x^n: A Recursive Approach

[Werg22]

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

 

[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.

 

[HallsofIvy]

How about using induction and the product rule?

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

 

[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.

 

[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]<br /> Then <br /> \frac{dx^{k+1}}{dx}= x^k+ x(kx^k)= (k+1)x^k<br /> <br /> By induction, then, \frac{dx^n}{dx}= nx^{n-1} for every positive integer n.<br /> <br /> 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.

 

[matt grime]

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).

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

Then the right side is simply a geometric series, so sum that up and you get na^(n-1).

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.

 

[JasonRox]

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.

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.

 

[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

 

[HallsofIvy]

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

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