Question

http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-a-definition-and-basic-rules/session-2-examples-of-derivatives/MIT18_10SCF10_Ses2d.pdf

Please help? I can't seem to wrap my head around the expansion of the binomial theorem. I know the formula but the written explanation for how n(Δx)x^(n-1) was arrived makes absolutely no sense to me.

Thanks! :)

Answer 1

First, here is a reference to the binomial theorem
http://en.wikipedia.org/wiki/Binomial_theorem#Statement_of_the_theorem

Perhaps if we find the derivative of f(x) = x^3, it will be clearer.

We use this definition:
\[ f'(x) = \lim_{h \rightarrow 0} \frac{ f(x+h) - f(x)}{(x+h) - x}\\
f'(x) = \lim_{h \rightarrow 0} \frac{ f(x+h) - f(x)}{h}
\]

with \( f(x)= x^3\):
\[ f'(x) = \lim_{h \rightarrow 0} \frac{ (x+h)^3 - x^3}{h}
\]

we can expand \( (x+h)^3\) using polynomial multiplication (a bit painful) or using the binomial theorem. Either way, we get
\[ (x+h)^3 = x^3 + 3x^2 h + 3 x h^2 + h^3 \]
notice that, except for the first two terms, all terms have *\(h^2\) or greater*

using the expanded expression in the definition:
\[ f'(x) = \lim_{h \rightarrow 0} \frac{ x^3 + 3x^2 h + 3 x h^2 + h^3 - x^3}{h} \]
the first and last term "cancel", so we have
\[ f'(x) = \lim_{h \rightarrow 0} \frac{ 3x^2 h + 3 x h^2 + h^3}{h} \]

Next, let's factor an "h" from each term:
\[ f'(x) = \lim_{h \rightarrow 0} \frac{ (3x^2 + 3 x h + h^2)h}{h} \\
=\lim_{h \rightarrow 0} \ (3x^2 + 3 x h + h^2) \frac{h}{h} \\
=\lim_{h \rightarrow 0} \ (3x^2 + 3 x h + h^2)
\]

Now take the limit as h approaches 0. Only the first term survives.
\[ f'(x) = 3x^2 \]

Answer 2

To make the argument more general, i.e. for exponent n, we note a few things:
1. when we expand (x+h)^n we will *always* get
\[x^n + n x^{n-1} h + \left(\begin{matrix}n \\ 2\end{matrix}\right) x^{n-2} h^2 + ... + \left(\begin{matrix}n \\ n\end{matrix}\right)h^n\]
2.) we will *always* subtract off \( x^n\), leaving
\[n x^{n-1} h + \left(\begin{matrix}n \\ 2\end{matrix}\right) x^{n-2} h^2 + ... + \left(\begin{matrix}n \\ n\end{matrix}\right)h^n\]

3.) we will always be able to factor out an "h" from each term, which will "cancel" with the "h" in the bottom of the derivative definition, leaving
\[ \lim_{h\rightarrow 0} n x^{n-1} + \left(\begin{matrix}n \\ 2\end{matrix}\right) x^{n-2} h + ... + \left(\begin{matrix}n \\ n\end{matrix}\right)h^{n-1}\]

notice that all terms, except for the first term, have an "h"
as we let h go to zero, only the first term survives, leaving
\[ f'(x) = n \ x^{n-1} \]

Answer 3

I am not sure exactly what is unclear to you. If it is specifically about the binomial theorem, this might help
https://www.khanacademy.org/math/algebra2/polynomial_and_rational/binomial_theorem/v/binomial-theorem