Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I encountered this in my calculus book:

$$f\;'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$


$$\begin{align*} (x + h)^n &= (x + h)(x + h)...(x + h)\\ &=x^n + nhx^{n-1}+ \text{stuff involving }h^2\text{ as factor} \end{align*}$$

I don't get where that $nhx^{n-1}$ and stuff involving $h^2$ as factor come from. A little help please?

share|cite|improve this question
Do you know binomial theorem? – Martin Sleziak Nov 12 '11 at 10:41
The $x^2$ are probably $x^n$. – Did Nov 12 '11 at 10:46
I corrected an $x^2$ to $x^n$ and took out a bunch of redundant material. – Brian M. Scott Nov 12 '11 at 10:50
Just a isn't necessary to pick an answer immediately. It's usually advisable to wait a day or so to see what people have to say. I'm not saying this because Paul's answer is bad, only because other answers may be better explanations for you. – jprete Nov 12 '11 at 13:57
@jprete thanks for your advice. – Andrew Nov 12 '11 at 14:19
up vote 8 down vote accepted

The binomial theorem gives another answer here : $$(x+h)^n = \sum_{k=0}^n \binom{n}{k} x^k h^{n-k}$$

By expanding the first terms, we obtain $$(x+h)^n = x^n + \binom{n}{1}hx^{n-1} + \binom{n}{2} x^{n-2} h^{2}+\cdots$$

In all the terms in the dots, the power of $h$ is greater than 2, so you can factorize by $h^2$. Moreover $\binom{n}{1} =n$ : $$(x+h)^n = x^n + n h x^{n-1} + h^2 \times ( \cdots ) $$

share|cite|improve this answer

Let us consider what happens when we try to expand $$ (x+h)^n = (x+h)(x+h)\cdots (x+h) .$$

To expand the RHS, we must pick either $x$ or $h$ from each bracket, and multiply them together to produce a term. The full expansion consists of the sum of the terms produced over all combinations of us picking $x$ or $h$ from each bracket.

What if we pick no $h$'s from any of the brackets? Then we pick $x$ every time, and the term produced is $x^{n}.$

Now, what if we pick $h$ exactly from the 1-st bracket? Then we must pick $x$ from the other $n-1$ brackets, so the term produced is $hx^{n-1}.$ Now if we pick $h$ exactly once, but from the 2nd bracket, same term is produced. We can pick our single $h$ from any of the $n$ brackets, so that means all the terms produced by picking exactly one $h$ sum to $nhx^{n-1}.$

Now, we've considered what happens if we pick 0 $h$'s (we get $x^n$) and 1 $h$ (we get $nhx^{n-1}$. Everything else must involve picking $h$ at least twice, and if we pick $h$ twice the term produced has at least a factor of $h^2$ in it, explaining the "stuff involving $h^2$ as a factor" term.

share|cite|improve this answer
That is the only way to explain the binomial theorem! (Well, maybe not the only way, but it's a far sight better than at least some teachers teach it :) ) – tom Nov 13 '11 at 11:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.