How to prove the derivatives of powers?

[Owen-]

Ok, I just had a lecture recently the attached picture is what it was about. I understand it all, however at the end it says we havnt proven it yet - just wondering how DO you prove it then?

Thanks,
Owen

 

[Owen-]

Hmm instead of just taking the linear part of the binomial expansion, why wasn't the full thing written down? Surely If the full formula for binomial was written out it would prove what we wanted?

the (x^n) parts would cancel then the only term without h in it after you divide across by h you would be left with what you wanted...

 

[arildno]

Well.
You would need to make an explicit error estimate to show that the rest term goes to zero sufficiently fast, but that is quite easy.

Another way of doing this is by assuming the truth of the product rule, and prove the general formula by induction:

Assume
$$f_{k}(x)=x^{k}, f_{k}^{'}(x)=k*x^{k-1}$$
Note that this formula is true for k=1.

For arbitrary "k", we wish to prove that the formula holds for "k+1" as well:
$$f_{k+1}(x)=x^{k+1}=x*f_{k}(x)\to{f}_{k+1}^{'}(x)=1*x^{k}+x*k*x^{k-1}=(k+1)*x^{k}$$

Therefore, we have proved that the formula holds for all powers k.

 

[Owen-]

So the only thing my lecturer needed to do there was to

show that the rest term goes to zero sufficiently fast

? And it would have been proven?

 

[arildno]

Yes, in an epsilon-deltawise manner.
Rather tedious and longwinded; that's why he didn't bother to do it.

 

[Owen-]

Ah - ok, thanks! Any chance you could find a link to this proof?

Frankly i don't even know what to look for :S

Thanks again,
Owen.

 

[hunt_mat]

Try this proof:
$$f'(a)=\lim_{x\rightarrow a}\frac{x^{n}-a^{n}}{x-a}$$
But $$x^{n}-a^{n}$$ has a factor of x-a, then we can factorise.
$$x^{n}-a^{n}=(x-a)(x^{n-1}+ax^{n-2}+\cdots +xa^{n-2}+a^{n-1})$$
Then from here the limit is easy to find and the result follows.

 

[HallsofIvy]

I don't believe that is the point they are making. They have, though some details have been jumped over, proved that the derivative of $x^n$ is $n x^{n-1}$ for n any positive integer. When they say "this is true in general", I think that they mean that it is true for n any number, not just a positive integer.

The most direct way to prove that the derivative of $x^a$ is $ax^{a- 1}$ for a any number is to use logarithms.

If $y= x^a$ then $y= e^{ln(x^a)}= e^{a ln(x)}$. Assuming that you already know that the derivative of $e^x$ is $e^x$, the derivative of $ln(x)$ is $1/x$, and the chain rule (which is why they "haven't proved it yet"), then we can say
$$\frac{dy}{dx}= \left(e^{aln(x)}\right)\left(\frac{a}{x}\right)$$

And, now, since $e^{a ln(x)}= e^{ln(x^a)}= x^a$, that says that
$$\frac{dy}{dx}= x^a\frac{a}{x}= ax^{a-1}$$.