Re: Derivative a*x^n

On Mon, 4 Feb 2008, carel wrote:

Is my proof new or just a variation of known stuff.

Alternative proof to the dirivative of y = a*x^n which is dy/dx =
a*n*x^(n-1)

First off, recall dcf(x)/dx = c.df(x)/dx, so we can toss a*.
Why all the variables? How about straight forward chaining?

D(x^n) = D(e^(log x^n) = D(e^(n.log x)) = e^(n.log x) * n.D(log x)
= nx^n * 1/x

Let e = the natural base 2.718281828.....

Now let x = e^u , so that y = a* e^(u*n) , then we have dx/du = e^u ,
so that du/dx = 1/e^u = 1/x

Note also that u = ln(x) , so the above is also a proof that the
derivative of ln(x) is 1/x

It is? How about some straight think?

1 = Dx = D(e^log x) = e^log x * D(log x) = x.D(log x)

So dy/dx = dy/du * du/dx = a*n*[e^(u*n)] * 1/x

So dy/dx = a*n*(x^n) *1/x = a*n*x^(n-1)

All immediate from De^x = e^x and the chain rule
De^f(x) = e^f(x) * Df(x).
.