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I am trying to understand why the expression $(x+h)^n$ will always have a term $n\times h\times x^{n-1}$ in this proof for the power rule and am not understanding what the proof in the image below means when it says 'take $x$ from all but one factor and h from the remaining factors', as well as 'selecting $h$ from at least two factors'. What is meant by taking and selecting $x$ and $h$ by factors? enter image description here

Sebastiano
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Princess Mia
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    Hint: binomial theorem. – Sean Roberson Jan 11 '23 at 05:17
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    When you multiply $(x+h)^n$ out, you select $x$ or $h$ from each of the $n$ $(x+h)$ terms. This gives $2^n$ monomials in $x,h$. Selecting all $x$s gives $x^n$. Selecting all $x$s except for one $h$ gives $n$ terms $h x^{n-1}$, $x h x^{n-2} ( = h x^{n-1})$, etc, etc. – copper.hat Jan 11 '23 at 05:24
  • thank you- so 'select' in this context means 'choose to be included in an expression multiplying everything selected together'? – Princess Mia Jan 11 '23 at 05:26
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    I am not sure what you are asking in the last comment. When you multiply out $(x+h)^n$ there are $2^n$ terms, as you can select/choose $x$ or $h$ from each of the $n$ terms. There will be $\binom{n}{k}$ entries of the form $x^k h^{n-k}$. – copper.hat Jan 11 '23 at 05:50

2 Answers2

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I think the answer to

What is meant by taking and selecting and ℎ by factors?

is, essentially, the distributive law. When you multiply out $$ (a+b)(c+d) = ac + ad + bc + bd $$ the four terms in the expansion correspond to pairing each of the two summands in $a+b$ with each of the two in $c+d$. When you expand the product of three binomials $(a+b)(c+d)(e+f)$ there will be $8$ terms. starting with $ace$.

When you expand $(x+h)^n$ there will be $2^n$ terms. Each will have some $x$'s and some $h$'s. Precisely $n$ of them will have one $h$ and $n-1$ $x$'s.

The binomial theorem (referred to in other answers and comments) tells you precisely how many of the terms are of each of the forms $x^kh^{n-k}$.

Ethan Bolker
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Hint: For $h \ne 0,$ if $f(x)=x^N,$ then using the Binomial Theorem, we get

$$\frac{f(x+h)-f(x)}{h} = \frac{(x+h)^N - x^N}{h} = \frac{\left(\sum_{n=0}^N \binom{N}n x^{N-n} h^n \right)-x^N}{h}=\sum_{n=1}^N \binom{N}n x^{N-n} h^{n-1}.$$ So when you take the limit as $h \rightarrow 0,$ which term in the above sum remains ?

Vivek Kaushik
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