# Proof of the derivative of $x^n$

Can someone please explain why $(x^n)'=n\cdot x^{n-1}$?

Sorry for not writing it in math characters, I'm new here.

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Note that if $x>0$, then $\ln(x^n)=n\ln(x)$. Taking the derivative of each side, using $(\ln(t))'=\dfrac{1}{t}$, and the chain rule on the left side,

$$\dfrac{1}{x^n}\cdot(x^n)' = n\cdot\dfrac{1}{x}.$$

Multiplying by $x^n$ on both sides yields $(x^n)'=nx^{n-1}$.

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Simple and clever. Thanks for teaching :) –  yonyon100 Oct 10 '13 at 9:32
Are u really allowed the presume knowledge about the logarithm? Besides, using the algebraic properties of the logarithm uses an approach to the exponential function via powers while the analytical knowledge of the derivative of the logarithm uses an approach of the exponential via differential equations. But these are on first view to different definitions not conform with each other. That is your argument is cyclic and therefore the proof seems wrong! –  Freeze_S Jun 27 at 23:05
I like that this generalizes to all $n\in\Bbb R-\{0\}$. Of course, you need to know that $x^n$ is differentiable in the first place to use the chain rule, but this approach hints at what the derivative ought to be. –  Bryan Jun 27 at 23:17
Thank you for the comment @Freeze_S. I don't understand all of your comment. The answer to whether it is circular is that it need not be, and it depends on what definition of the logarithm you are using. For example, the natural logarithm can be defined to be $\ln(x) = \int_1^x \dfrac{1}{t}\,dt$. Using this definition (or others) one can prove the formula for the derivative of $\ln$ and the formula for $\ln(x^n)$ without needing to use the derivative of $x^n$. In any case, I don't recommend this as a first introduction to the power rule. –  Jonas Meyer Jun 28 at 5:54
@Freeze_S: It is straightforward to show that $\int_1^a \frac1t\,dt=\int_b^{ab}\frac1t\,dt$ for all positive $a$ and $b$, from which it follows that $\ln(ab)=\ln(a)+\ln(b)$. From this it one can play around to show that $\ln(a^n)=n\ln(a)$ for positive $a$ and rational $n$. To get the case of irrational $n$ requires a continuity argument. –  Jonas Meyer Jun 28 at 16:52

For any positive integer $n$, consider the following algebraic identity: $$\frac{x^n - x_0^n}{x-x_0} = x^{n-1} + x^{n-2}\cdot x_0 + \dots + x\cdot x_0^{n-2} + x_0^{n-1}$$ When $x \to x_0$, the limit is $n\cdot x_0^{n-1}$.

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I love this answer –  Learner Oct 10 '13 at 7:36
I must say that after some thought put into it, your answer is the most rational and understandable –  yonyon100 Oct 11 '13 at 5:41

For $x, h \in \mathbb R$, we have by the binomial theorem $$(x+h)^n = \sum_k \binom nk x^{n-k}h^k$$ Hence \begin{align*} \frac{(x+h)^n - x^n}h &= \frac 1h \sum_{k=1}^n \binom nk x^{n-k}h^{k}\\ &= nx^{n-1} + h \cdot \sum_{k=2}^n \binom nk x^{n-k}h^{k-2}\\ &\to nx^{n-1}, \quad h \to 0 \end{align*}

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I like this answer. I would add that it can generalize to $n$ other than positive integers by rewriting as $x^n\dfrac{1}{h}((1+\frac{h}{x})^n-1)$ and using the binomial series for $(1+\frac{h}{x})^n$ (which is perhaps what Nicky Hekster's hint is suggesting doing). –  Jonas Meyer Oct 10 '13 at 7:57
Why the downvote. Did I overlook something? –  martini Oct 10 '13 at 11:58
@Jonas, I was hinting at Martini's proof! –  Nicky Hekster Oct 11 '13 at 12:07

Hint: use the definition of the derivative and Newton's binomial formula.

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So you know $f' = \lim \frac{ \Delta f}{\Delta x}$. Now, suppose we are given $f$ and $g$, then we want $(fg)'$,

$$(fg)' = \lim_{\Delta x \to 0 } \frac{ f(x + \Delta x)g(x + \Delta x) - f(x)g(x)}{\Delta x} = \lim_{\Delta x \to 0 } \frac{ f(x + \Delta x)g(x + \Delta x) + f(x)g(x + \Delta x) - f(x)g(x + \Delta x) - f(x)g(x)}{\Delta x} = \lim_{\Delta x \to 0 } \frac{ g(x + \Delta x)[f(x + \Delta x) - f(x) ] + f(x)[g( x + \Delta x) - g(x)]}{\Delta x} = f'g + fg'$$

Now, since we have shown this trick, we can use it to show $(x^n)' = nx^{n-1}$ If $n = 1$, then $(x)' = 1 = 1x^{0}$. Suppose $(x^n)' = nx^{n-1}$ is for true for $n$, then

$$(x^{n+1})' = [(x^n)(x)]' = nx^{n-1} x + x^n = (n+1)x^n$$

The problem is now solved by induction. bye bye

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This works if $n$ is a positive integer. –  Jonas Meyer Oct 10 '13 at 7:16
Z is symmetric . –  Learner Oct 10 '13 at 7:18
I don't know what that means. –  Jonas Meyer Oct 10 '13 at 7:18
$x^{-n} = (x^{-1})^n$ now apply the trick above –  Learner Oct 10 '13 at 7:20
So you are suggesting using induction to prove the negative integer case? Sure, that would work. Or one could use the quotient rule and the case you already showed, or use the product rule applied to the equation $x^n\cdot x^{-n} = 1$. Then you would have the case of integers. The notation "$n$" might hint that only integers are desired. –  Jonas Meyer Oct 10 '13 at 7:23

This works when $n\in \mathbb{Z}$ and you know the basic properties of derivative.

Let $n\in \mathbb{N}$ we proceed by induction. When $n=0$, so $x^n$ is a constant and then $(x^n)'=0= nx^{n-1}$. Suppose we have already proven the assertion for $n\ge 0$, then $(x^{n+1})'=(x\cdot x^n)'=x'x^n+(x^n)'x=x^n+nx^{n-1}x=x^n+nx^n=(n+1)x^n$.

If $n<0$ and $x\not =0$, then $(x^n)=(1/x^m)$ where $m=-n$ so $m\in \mathbb{N}$. Then $(1/x^m)'=-(mx^{m-1})/x^{2m}=-mx^{-m-1}$ and so $nx^{n-1}$.

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Do you know the Product Rule? If so, there's a cleaner method than the binomial formula (if not, the binomial formula is your best bet).

You have $x^n = \underbrace{x \cdot x \cdot \dots \cdot x \cdot x}_{n \, \text{ times}}$. The Product Rule says that the derivative is equal to:

$$\underbrace{(x')\cdot\underbrace{x \cdot x \cdot \dots \cdot x \cdot x}_{n-1 \,\text{ times}} + x \cdot (x') \cdot \underbrace{x \cdot x \cdot \dots \cdot x \cdot x}_{n-2 \, \text{ times}} + \dots + \underbrace{x \cdot x \cdot \dots \cdot x \cdot x}_{n-1 \, \text{ times}}\cdot (x')}_{n \,\text{ total products added together}}$$

We know that $(x') = 1$. So the above sum is equal to:

$$\underbrace{\underbrace{x \cdot x \cdot \dots \cdot x \cdot x}_{n-1 \,\text{ times}} + \dots + \underbrace{x \cdot x \cdot \dots \cdot x \cdot x}_{n-1 \, \text{ times}}}_{n \,\text{ times}}$$

Which is clearly equal to $nx^{n-1}$.

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This works if $n$ is a positive integer. (Newton's binomial formula is more general.) –  Jonas Meyer Oct 10 '13 at 7:23

I somewhat like this short derivation that exploits the power of the binomial theorem: \begin{align*} \boxed{f(x)=x^n}\quad f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}&=\lim_{h\to0}\dfrac{x^n+nx^{n-1}h+\cdots+h^n-x^n}{h}\\ &=\lim_{h\to0}\dfrac{nx^{n-1}h+\mathcal O(h^2)} {h}\\ &=\lim_{h\to0}\dfrac{nx^{n-1}h}h+\dfrac{\mathcal O(h^2)}h \\ &=nx^{n-1}+\lim_{h\to0}\mathcal{O}(h) \\&=nx^{n-1} \quad {\tiny\blacksquare} \end{align*}

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