# Expansion of $(x+h)^n$

I was reading the proof for the power rule for derivatives, and I came across this step:

$\displaystyle \frac{\left({x + h}\right)^n - x^n} h = \displaystyle \frac{x^n} h \left({\left({1 + \frac h x}\right)^n - 1}\right)$

I was wondering if someone could elaborate on the inner workings of this step? They seem to have been able to expand $(x+h)^n$ without much trouble, and without the need for a sigma sum.

To clarify, I understand factorization, and how they extracted $\displaystyle \frac{x^n}h$, but everything before that is a haze.

• Just factor out $x^n$ – Fakemistake Nov 12 '17 at 18:47

The basic idea is to pull out a factor of $x^n$ in the numerator. The rest is fairly straight-forward algebraic computation. \begin{align} \frac{(x+h)^n - x^n}{h} &= \frac{\left( x + \frac{x}{x} h \right)^n - x^n }{h} && (\text{multiply by 1}) \\ &= \frac{\left( x \left(1+\frac{h}{x}\right) \right)^n - x^n}{h} &&(\text{factor out $x$}) \\ &= \frac{x^n \left(1+\frac{h}{x}\right)^n - x^n}{h} && (\text{exponentiation rules}) \\ &= \frac{x^n \left( \left(1+\frac{h}{x}\right)^n - 1\right)}{h} && (\text{factor out $x^n$}) \\ &= \frac{x^n}{h} \left( \left(1+\frac{h}{x}\right)^n - 1\right). && (\text{simplify}) \end{align}
Big Hint:\begin{align*}\left(x+h\right)^n-x^n & =\left[x\left(1+\frac hx\right)\right]^n-x^n\\ & =x^n\left(1+\frac hx\right)^n-x^n\end{align*}