Proof of the derivative of $x^n$

Question

I am proving $(x^n)'=nx^{n-1}$ by the definition of the derivative:

\begin{align} (x^n)'&=\lim_{h \to 0} {(x+h)^n-x^n\over h}\\ &=\lim_{h \to 0} {x^n+nx^{n-1}h+{n(n-1)\over 2}x^{n-2}h^2+\cdots+h^n-x^n\over h} \\ &=\lim_{h \to 0} \left[ nx^{n-1}+{n(n-1)\over 2}x^{n-2}h+\cdots+h^{n-1} \right] \end{align}

Because polynomial is continuous for every $x$, we can conclude that $\lim_{x_0\to 0}(x_0)^n=0$. Therefore $$\lim_{h \to 0} \left[ nx^{n-1}+{n(n-1)\over 2}x^{n-2}h+\dots+h^{n-1} \right]= nx^{n-1}$$

Is this proof valid?

Answer 1

It is correct, but too many ellipses tend to obscure a bit the proof. And ellipses are not very rigurous, if you ask me.

I'd rewrite it using this:

For $n\ge 2$ there exists some polynomial $P(x,h)$ such that $$(x+h)^n=x^n+nhx^{n-1}+h^2P(x,h)$$

Answer 2

The proof in OP is correct if $n$ is a positive integer. For a generic real exponent $a$ we can start from the derivative of the exponential function $y=e^x \rightarrow y'=e^x$.

From the inverse function differentiation rule we find $y=\log x \rightarrow y'=\dfrac{1}{x}$ and (using the chain rule):

$$ y=x^a =e^{a \log x} \rightarrow y'=e^{a \log x} (a\log x)'=e^{a \log x} \dfrac{a}{x}=ax^{a-1} $$

Answer 3

Quote of the Question to Make it Easier to Reference in the Answer

I am proving $(x^n)'=nx^{n-1}$ by the definition of the derivative: $$ \begin{align} (x^n)'&=\lim_{h \to 0} {(x+h)^n-x^n\over h}\\ &=\lim_{h \to 0} {x^n+nx^{n-1}h+{n(n-1)\over 2}x^{n-2}h^2+\cdots+h^n-x^n\over h} \\ &=\lim_{h \to 0} \left[ nx^{n-1}+{n(n-1)\over 2}x^{n-2}h+\cdots+h^{n-1} \right] \end{align} $$ Because polynomial is continuous for every $x$, we can conclude that $\lim_{x_0\to 0}(x_0)^n=0$. Therefore $$\lim_{h \to 0} \left[ nx^{n-1}+{n(n-1)\over 2}x^{n-2}h+\dots+h^{n-1} \right]= nx^{n-1}$$ Is this proof valid?

Steps Towards the Answer

By application of the Binomial Theorem for example for $(x+y)^4$: The coefficient a in the term of $ax^by^c$ is known as the binomial coefficient ${\displaystyle {\tbinom {n}{b}}}$ or ${\displaystyle {\tbinom {n}{c}}}$ (the two have the same value). These coefficients for varying n and b can be arranged to form Pascal's triangle.

The derivative of a constant $C$ $\left(n=0\right)$ is zero since $\displaystyle \lim_{h \to 0}\frac{C-C}{h}=\displaystyle \lim_{h \to 0}\frac{0}{h}=0$. Now when $\displaystyle \lim_{h\to 0}$ then for the case $n=1$: $$x^{'}=\lim_{h \to 0}\frac{x+h-x}{h}=\lim_{h \to 0} \frac{h}{h}=1$$ Now consider $n \ge 2$ so that:

$$\lim_{h \to 0}\frac{(x+h)^n-x^n}{h}= \lim_{h \to 0}\frac{ {\tbinom {n}{n}} \,x^n + h \, {\tbinom {n}{n-1}} \,x^{n-1}+ {\tbinom {n}{n-2}}\,x^{n-2}*h^2 + \, ...\text{Higher Order Terms of $h$} -x^n}{h} $$ Since the binomial coefficient ${\tbinom {n}{n}}=1$ and the binomial coefficient ${\tbinom {n}{n-1}}=n$ and $x^2-x^2=0$, then: $$ \lim_{h \to 0}\frac{(x+h)^n-x^n}{h}= \lim_{h \to 0}\frac{ n \, h \, x^{n-1}+ {\tbinom {n}{n-2}}\,x^{n-2}*h^2 + \, ...\text{Higher Order Terms of $h$} }{h} $$ Taking the limit $\displaystyle \lim_{h \to 0}$ where $\displaystyle \lim_{h \to 0} h=0 $ and Higher Order Terms of h $\displaystyle \lim_{h \to 0} \, \text{and (Higher Order Terms of h) } \to 0 \, \text{ so: }$ $$ \begin{align} \lim_{h \to 0}\frac{(x+h)^n-x^n}{h}&= \lim_{h \to 0}\frac{ n \, h \, x^{n-1}+ {\tbinom {n}{n-2}}\,x^{n-2}*h^2 + \, ...\text{Higher Order Terms of $h$} }{h} \\ &=\lim_{h \to 0} n \, x^{n-1}+ {\tbinom {n}{n-2}}\,x^{n-2}*h + \, ...\text{Higher Order Terms of $h$} \\ &=\lim_{h \to 0} n \, x^{n-1}=n \, x^{n-1} \end{align} $$ So, $$\boxed{ \left(x^n\right)^{'}=\lim_{h \to 0}\frac{(x+h)^n-x^n}{h}=n \, x^{n-1} }$$ And the proof is complete. Aside from a few formalities and additional details, these were basically the same steps as summarized with the question's proof. So the summary proof in the question is complete and correct.

Answer 4

Another proof of

$$(x^n)'=nx^{n-1}, \forall n \in \mathbb{N}$$

can be implemented with induction on $n$ using the product rule

$$(f \cdot g)' = f'g+fg'$$

and the trivial identity derivative

$$(x)'=1$$

Specifically:

1. For $n=0$, $(x^0)'=(1)'=0=0 \cdot x^{-1}$ holds
2. For $n=1$, $(x^1)'=1=1 \cdot x^{0}$ holds
3. Assume that $(x^k)'=kx^{k-1}$ for some $k \in \mathbb{N}$ holds
4. For $k+1$ using the product rule we have:
   $$(x^{k+1})' = (x \cdot x^k)'$$ $$(x^{k+1})' = x \cdot kx^{k-1}+1 \cdot x^k$$ $$(x^{k+1})' = (k+1)x^{k}$$
5. Thus $(x^n)'=nx^{n-1}, \forall n \in \mathbb{N}$ holds. QED

Furthermore:

1. $(x^{-n})'=-nx^{-n-1}, \forall n \in \mathbb{N}$ can be proved with chain rule on $((\frac{1}{x})^n)'$
2. $(x^{1/n})'=\frac{1}{n}x^{1/n-1}, \forall n \in \mathbb{N}$ can be proved with inverse rule on $f^{-1}(x)=x^n \implies f'=1/{f^{-1}}'$
3. $(x^{n/m})'=\frac{n}{m}x^{n/m-1}, \forall n,m \in \mathbb{N}$ can be proved with chain rule on $((x^{1/m})^{n})'$
4. Similarly for $(x^{-n/m})'=-\frac{n}{m}x^{-n/m-1}, \forall n,m \in \mathbb{N}$
5. Thus, $(x^q)'=qx^{q-1}, \forall q=\frac{n}{m} \in \mathbb{Q}$. QED

Answer 5

This is correct if $n\neq 0$: $\lim_{x_0\to 0}(x_0)^n=0$. Otherwise, it is $1$. You should have said that $\lim_{h\to 0}h^k=0$ for $k\geq 1$.

And your prove, as it is written, is valid for $n\in\Bbb N$. Here is a cute implicit differentiation proof for all $n\in\Bbb R$: $y=x^n\implies \ln y=n\ln x\implies \frac{y'}y=\frac nx\implies y'=nx^{n-1},$ since we know that $(\ln x)'=\frac1x.$