show formula f'(x) = nxn^-1 comes from 2nd term in binomial-

happyman

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Show that the formula for the derivative of a power, namely, f ' (x) = nxn^ -1, comes from the second term in the binomial series expansion of (x+ delta x)^n
 
Re: derivative

suppose f(x) = x^n, n is an integer greater than 0

note that I'm using h, which is synonymous with delta x, but easier to write.

then f'(x) = lim [h --> 0] [f(x+h) - f(x)] / h

= lim [h-->0] [(x+h)^n - x^n] / h

= lim [h-->0] {[x^n + nx^(n-1)h + n(n-1)/2 * x^(n-2)h^2 + ... + nxh^(n-1) + h^n] - x^n} /h

- this comes from rewriting (x+h)^n according to the definition of binomial expansion

the first term x^n will cancel out with the x^n you subtract at the end. Then factor out an h everwhere on top and cancel out top and bottom. Every term on top will still have an h except the 2nd term of the binomial expansion: nx^(n-1)h (because the h has been factored out). Direct substitute h = 0 and every term becomes 0 except nx^(n-1).

Can you figure out the proof for the case where n is an integer less than 0? It involves the quotient rule.

Also you can prove it for the case where n is any real number, but the proof involves using logarithms, not binomial expansion.
 
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