Power Rule for derivatives

**Krizalid** · October 17th 2008, 11:03 AM

Let $f(x)=x^n$ and prove that its derivative is given by $f'(x)=n\cdot x^{n-1},\,\forall\,n\in\mathbb R.$

**Krizalid** · October 17th 2008, 11:14 AM

There's a problem with that solution 'cause binomial theorem works with natural numbers, hence, when computing the limit, you assume that $n$ is natural since $n$ is a real number.

**Krizalid** · October 17th 2008, 11:23 AM

Yes, I know, but in that case we state that such derivative only works for natural numbers.

**Moo** · October 17th 2008, 01:44 PM

Hello,

Originally Posted by Krizalid

Yes, I know, but in that case we state that such derivative only works for natural numbers.

Generalized Newton's binomial formula :

$(x+y)^n=\sum_{k=0}^\infty {n \choose k} x^k y^{n-k}$

where ${n \choose k}=\frac{r(r-1)\dots(r-k+1)}{k!}$ and $\forall n \in \mathbb{R}$

_______________________________________
A way of doing it is to remember another formula for the derivative :
(assuming it is differentiable so at least $x_0 \neq 0$ )

$f'(x_0)=\lim_{x \to x_0} ~ \frac{x^n-x_0^n}{x-x_0}$

$\frac{x^n-x_0^n}{x-x_0}=\frac{x_0^n}{x_0} \times \frac{\left(\frac{x}{x_0}\right)^n-1}{\frac{x}{x_0}-1}=x_0^{n-1} \times \sum_{k=0}^{n-1} \left(\frac{x}{x_0}\right)^k$

$\lim_{x \to x_0} ~ x_0^{n-1} \times \sum_{k=0}^{n-1} \left(\frac{x}{x_0}\right)^k=x_0^{n-1} \times \sum_{k=0}^{n-1} 1^k=x_0^{n-1} \times n \quad \square$

**Krizalid** · October 18th 2008, 07:09 AM

This is my approach:

$\begin{aligned} \frac{(x+\Delta x)^{n}-x^{n}}{\Delta x}&=\frac{e^{n\ln (x+\Delta x)}-e^{n\ln x}}{\Delta x} \\ & =x^{n}\cdot \frac{e^{n\ln \left( 1+\frac{\Delta x}{x} \right)}-1}{\Delta x} \\ & =x^{n}\cdot \frac{e^{n\ln \left( 1+\frac{\Delta x}{x} \right)}-1}{n\ln \left( 1+\dfrac{\Delta x}{x} \right)}\cdot \frac{n\ln \left( 1+\dfrac{\Delta x}{x} \right)}{\Delta x}, \end{aligned}$

hence,

$\frac{(x+\Delta x)^{n}-x^{n}}{\Delta x}=x^{n}\cdot \frac{e^{n\ln \left( 1+\frac{\Delta x}{x} \right)}-1}{n\ln \left( 1+\dfrac{\Delta x}{x} \right)}\cdot \frac{n\ln \left( 1+\dfrac{\Delta x}{x} \right)}{\dfrac{\Delta x}{x}}\cdot \frac{1}{x}\to nx^{n-1}$ as $\Delta x\to0.\quad\blacksquare$

**Moo** · October 18th 2008, 07:36 AM

Originally Posted by Krizalid

This is my approach:

$\begin{aligned} \frac{(x+\Delta x)^{n}-x^{n}}{\Delta x}&=\frac{e^{n\ln (x+\Delta x)}-e^{n\ln x}}{\Delta x} \\ & =x^{n}\cdot \frac{e^{n\ln \left( 1+\frac{\Delta x}{x} \right)}-1}{\Delta x} \\ & =x^{n}\cdot \frac{e^{n\ln \left( 1+\frac{\Delta x}{x} \right)}-1}{n\ln \left( 1+\dfrac{\Delta x}{x} \right)}\cdot \frac{n\ln \left( 1+\dfrac{\Delta x}{x} \right)}{\Delta x}, \end{aligned}$

hence,

$\frac{(x+\Delta x)^{n}-x^{n}}{\Delta x}=x^{n}\cdot \frac{e^{n\ln \left( 1+\frac{\Delta x}{x} \right)}-1}{n\ln \left( 1+\dfrac{\Delta x}{x} \right)}\cdot \frac{n\ln \left( 1+\dfrac{\Delta x}{x} \right)}{\dfrac{\Delta x}{x}}\cdot \frac{1}{x}\to nx^{n-1}$ as $\Delta x\to0.\quad\blacksquare$

Pardon me for that, but I do not find it elegant

First of all, why don't you directly factor by $x^n$ in the beginning instead of factorizing with exponentials ?

Also, what if $x+\Delta x < 0$ ?

Plus, you ought to know the limits $\lim_{x \to 0} ~ \frac{\ln(1+x)}{x}=1$ and $\lim_{x \to 0} ~ \frac{e^x-1}{x}=1$ , which are proved using the definitions of derivative numbers (if not "advanced" calculus...)

**Krizalid** · October 18th 2008, 07:45 AM

Originally Posted by Moo

Pardon me for that, but I do not find it elegant

Who said it was elegant?

Originally Posted by Moo

First of all, why don't you directly factor by $x^n$ in the beginning instead of factorizing with exponentials ?

To use known limits.

Originally Posted by Moo

Also, what if $x+\Delta x < 0$ ?

Why should I consider that case?

Originally Posted by Moo

Plus, you ought to know the limits $\lim_{x \to 0} ~ \frac{\ln(1+x)}{x}=1$ and $\lim_{x \to 0} ~ \frac{e^x-1}{x}=1$ , which are proved using the definitions of derivative numbers (if not "advanced" calculus...)

By just knowing that $\lim_{x\to\infty}\left(1+\frac1x\right)^x=e$ those limits are easy to prove.

**Moo** · October 18th 2008, 07:55 AM

Originally Posted by Krizalid

Who said it was elegant?

Just giving my point.

To use known limits.

You factorize later, after changing into exponentials, while it would have been exactly the same if you did it from the beginning.

Why should I consider that case?

$\ln(x+\Delta x)$ when $x + \Delta x< 0$ ?

By just knowing that $\lim_{x\to\infty}\left(1+\frac1x\right)^x=e$ those limits are easy to prove.

What's the interest of asking to prove a simple formula if it has to use nested limits ?

Well, I'm no one to tell you this, so just ignore it if you want.

**ThePerfectHacker** · October 18th 2008, 05:17 PM

Krazalid is correct to say limit definition is no good if $n$ is not an integer.

And generalized binomial theorem is faulty since you implicity use the power rule in its derivation.

The standard way is to notice that by definition $x^r = \exp(r\ln x)$ for $x>0$ .
Here $\exp$ is the inverse function for $\ln t = \smallint_1^t \tfrac{du}{u}$ .

If we are familar with the basic rules for $\ln$ and $\exp$ just apply the chain rule and get your answer.

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