How to prove that the derivative of x^n=nx^(n-1)

**citcat** · September 1st 2012, 11:19 AM

How do I simplify the limit as h approaches 0 of ((x+h)^n-x^n)/h to prove that the derivative of x^n=nx^(n-1)?
Thanks!

**Plato** · September 1st 2012, 11:34 AM

Originally Posted by citcat

How do I simplify the limit as h approaches 0 of ((x+h)^n-x^n)/h to prove that the derivative of x^n=nx^(n-1)?

${\left( {x + h} \right)^n} - {x^n} = \sum\limits_{k = 1}^n {\binom{n}{k}{x^{n - k}h^k}}$

$\frac{{\sum\limits_{k = 1}^n {\binom{n}{k}{x^{n - k}}{h^k}} }}{h} = \sum\limits_{k = 1}^n {\binom{n}{k}{x^{n - k}}{h^{k - 1}}}$

**emakarov** · September 1st 2012, 11:39 AM

Another way. Let y = x + h; then y - x = h. We have $y^n - x^n=(y-x)(y^{n-1}+y^{n-2}x+\dots+yx^{n-2}+x^{n-1})$ . Now $y^{n-1}+\dots+x^{n-1}\to nx^{n-1}$ as $y\to x$ .

**Deveno** · September 1st 2012, 02:19 PM

an "informal proof"

[(x+h)ⁿ - xⁿ]/h =

(xⁿ + nx^n-1h + (other stuff with h² terms) - xⁿ)/h =

(nx^n-1h)/h + h²(...stuff...)/h =

nx^n-1 + h(...stuff...).

no matter what the "...stuff..." is, it is clearly not infinite at any particular x, so we have:

$\lim_{h \to 0} \frac{(x+h)^n - x^n}{h} = nx^{n-1} + (0)(...\text{limit of stuff}...) = nx^{n-1}$ .

Plato's post makes this argument rigorous, but this gives you "the general idea".

or, you can use induction:

base case: n = 1

then $\lim_{h \to 0} \frac{(x+h) - x}{h} = \lim_{h \to 0} \frac{h}{h} = \lim_{h \to 0} 1 = 1 = x^0$ .

assume that for n = k-1:

$\lim_{h \to 0} \frac{(x+h)^{k-1} - x^k}{h} = (k-1)x^{k-2}$ .

then:

$\lim_{h \to 0} \frac{(x+h)^k - x^k}{h} = \lim_{h \to 0} \frac{(x+h)(x+h)^{k-1} - (x+h)x^{k-1} + (x+h)x^{k-1} - x(x^{k-1})}{h}$

$= \lim_{h \to 0}\left[ (x+h)\left(\frac{(x+h)^{k-1} - x^{k-1}}{h}\right)\right] + \lim_{h \to 0}\left[x^{k-1}\left(\frac{(x+h) - x}{h}\right)\right]$

$= \left(\lim_{h \to 0} (x+h) \right) \left(\lim_{h \to 0} \frac{(x+h)^{k-1} - x^{k-1}}{h}\right) + \left(\lim_{h \to 0} x^{k-1} \right) \left( \lim_{h \to 0} \frac{(x + h) - x}{h} \right)$

$= \left(\lim_{h \to 0} (x+h) \right)((k-1)x^{k-2}) + \left(\lim_{h \to 0} x^{k-1} \right)(1)$ , by our induction hypothesis, and the base case,

$= (x)((k-1)x^{k-2}) + x^{k-1} = (k-1)x^{k-1} + x^{k-1} = (k-1+1)x^{k-1} = kx^{k-1}$ , and we are done.

**TheSaviour** · September 1st 2012, 03:52 PM

Originally Posted by emakarov

Another way. Let y = x + h; then y - x = h. We have $y^n - x^n=(y-x)(y^{n-1}+y^{n-2}x+\dots+yx^{n-2}+x^{n-1})$ . Now $y^{n-1}+\dots+x^{n-1}\to nx^{n-1}$ as $y\to x$ .

Is the above proof valid if, for example, $n = \sqrt{2}$ ?

**HallsofIvy** · September 1st 2012, 04:40 PM

No, it isn't.

Another proof, for n an integer, would be inductive: if n= 0, then $x^0= 1$ is a constant so its derivative is $0= 0x^{0-1}$ . Assume that, for some k, the derivative of $x^k$ is $kx^{k-1}$ . Then we can differentiate $x^{k+1}= x(x^k)$ by the product rule: the derivative is $(x)'(x^k)+ (x)(x^k)'= 1(x^k)+ x(kx^{k-1})= x^k+ x(kx^{k-1}= (k+1)x^k$ .

For k not an integer, we can use "logarithmic differentiation". If $y= x^k$ , then $ln(y)= ln(x^k)= kln(x)$ . Now, we need the fact that the derivative of ln(x) is 1/x (which why we typically start with integer powers of x). The derivative of the left side is $\frac{1}{y}\frac{dy}{dx}$ and the derivative of the right side is $\frac{k}{x}$ . Since $y= x^k$ , we have $\frac{1}{x^k}\frac{dy}{dx}= \frac{k}{x}$ . From that, we have $\frac{dy}{dx}= \frac{k}{x}x^k= kx^{k-1}$ .

**Prove It** · September 1st 2012, 08:26 PM

Originally Posted by citcat

How do I simplify the limit as h approaches 0 of ((x+h)^n-x^n)/h to prove that the derivative of x^n=nx^(n-1)?
Thanks!

You can use the Binomial Expansion to prove the result for positive integer values of n. Then you can use the chain rule to extend this result to other values of n.

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