To expand you will need to use the binomial theorem.
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
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
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?