## chand101 3 years ago This question may sound stupid but how does the binomial theorem relate to d/dx of x^n?

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1. DOLFIN

BECAUSE AFTER THER SECOND TERM DX IS TO THE POWER OF TWO OR HIGHER , THEREFORE , IT CAN'T BE DIVIDED BY DX , IN THE DENOMINATOR , SO WHEN YOU TAKE THE LIMIT OF DX TO CERO EVERY TERM IS ELIMINATED BUT THE SECOND ONE BECAUSE IT DOESN'T CONTAIN DX '

2. calculusfunctions

Binomial theorem can be used to prove that$\frac{ d(x ^{n}) }{ dx }=nx ^{n -1}$

3. wdngyre

Because when you are finding d/dx using the limit method (delta x going to zero), the binomial theorem allows you to simplify your difference quotient by giving you this algebraic expression: $(x+Deltax)^{n}=x ^{n}+n(x ^{n-1}Deltax)$ (The rest of the terms disappear when delta x goes to zero, so you don't need to worry about them in this particular problem.) You can now simplify: $\lim_{x \rightarrow 0}\frac{ (x+Deltax)^{n}-x ^{n} }{Deltax}=nx ^{n-1}$

4. pasta

wdngre has it .the binomial theorem helps expand the function for better/easier differenciation.in a way it RELATES

5. merckens

I think it's probably worthwhile to understand why the rest of the binomial formula is junk when you're finding: $\frac{ d }{ dx }(x ^{n})$So we have the sum: $\left( x + \Delta(x) \right)^{n} = x ^{n} + nx ^{n-1}\Delta(x) + junk$Now let's consider the first item in the junk. It's going to be: $some\_number*x^{n-2}\Delta(x)^2$(Note: some_number is known as the "binomial coefficient".) And the next item in the sequence: $some\_number*x^{n-3}\Delta(x)^3$And so on until you get to the last two items: $nx\Delta(x)^{n-1}+\Delta(x)^n$Now, when in the lecture Professor Jerison makes the key cancellation and you divide through by delta(x), consider what happens to all of the products in the junk. They all include delta(x) raised to some power greater than 1 (from 2 all the way up to n). When you divide through by delta(x), you're only eliminating one of those delta(x) values, so you have at least one delta(x) remaining in the product. For example: $some\_number*x^{n-2}\Delta(x)^2\rightarrow some\_number*x^{n-2}\Delta(x)$ You're always going to have at least one. So when delta(x) approaches 0, you're always going to be multiplying those products by 0. That's why they're junk. Because all of the products in the junk approach 0 as delta(x) approaches 0. They all just cancel out.