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.
 A: 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 $(x^n)'=nx^{n-1}$.
A: 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*}
A: Hint: use the definition of the derivative and Newton's binomial formula.
A: 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 $n\cdot x_0^{n-1}$.
A: 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
A: 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 $nx^{n-1}$.
A: I somewhat like this short derivation that exploits the power of the binomial theorem: $$\begin{align*}
\boxed{f(x)=x^n}\quad f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}&=\lim_{h\to0}\dfrac{x^n+nx^{n-1}h+\cdots+h^n-x^n}{h}\\ &=\lim_{h\to0}\dfrac{nx^{n-1}h+\mathcal O(h^2)} {h}\\
&=\lim_{h\to0}\dfrac{nx^{n-1}h}h+\dfrac{\mathcal O(h^2)}h \\
&=nx^{n-1}+\lim_{h\to0}\mathcal{O}(h)
\\&=nx^{n-1} \quad {\tiny\blacksquare}
\end{align*}$$
A: 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 $nx^{n-1}$.
