proof: derivative for (ax + b)^n

So the title says it all:

Write down the expansion of (ax + b)^n where n is a positive integer. Differentiate the result with respect to x. Show that the derivative is na((ax + b)^(n-1))

So I'm not even sure what I'm looking for here, or how much help I'll get here. This obviously isn't a test question or a question that is going to be corrected, and what would actually help me most is a proof or link to a proof so that I can see where I'm wrong. I'm not even sure I'm wrong. I've just come to a point where its hard to wrap my head around whats going on.

So I've got to the point where I have expanded (ax + b)^n, I have differentiated it and I have (a) as a factor of (ax + b)^(n-1), except with (n) as a factor of the first term (ax)^(n-1). Now there's no way I can post all of my work, which includes Pascal notation or whatever the conventional term is.

Anyway I'd be thankful if someone having the appropriate software could provide me with a proof so I can see where I've gone wrong.

Thanks

Re: proof: derivative for (ax + b)^n

To expand you will need to use the binomial theorem.

Re: proof: derivative for (ax + b)^n

Or you can use induction to prove that the derivative of is .

If n= 1, the derivative of [tex](ax+ b)^1= ax+ b[tex] is .

Assume that, for some k, the derivative of is .

Then and, by the product rule, the derivative is

Re: proof: derivative for (ax + b)^n

Re: proof: derivative for (ax + b)^n

alright thanks so much for your time, this is exactly what i was looking for. however, why did you expand (ax + b)^n from the highest powers of b to the highest powers of ax, while keeping the pascal terms in their original order? i understand that pascal sequence is symmetrical, but still I was wondering why you found this a more useful method.

Also I think you made a mistake when you started developing d((ax + b)^n)/dx, you omitted b in the coefficient of x starting from the second to last term, or is this on purpose?

Re: proof: derivative for (ax + b)^n

still, the first answer submitted has been the most helpful so far and the closest in form and notation to what i'm looking for, but as abovementioned i think there was a mistake in its development, ca

Re: proof: derivative for (ax + b)^n

Quote:

Originally Posted by

**furor celtica** still, the first answer submitted has been the most helpful so far and the closest in form and notation to what i'm looking for, but as abovementioned i think there was a mistake in its development, ca

You are right. I made a mistake. You might call that a typo.

Here is the corrected version of my reply.