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 $\displaystyle (ax+b)^n$ you will need to use the binomial theorem.

$\displaystyle (ax+b)^n=\sum_{i=0}^{n}\begin{pmatrix}n \\ i\end{pmatrix}(ax)^{i}b^{(n-i)}$

$\displaystyle \begin{align*}\implies (ax+b)^n &=\begin{pmatrix}n \\0\end{pmatrix}b^n+\begin{pmatrix}n \\1\end{pmatrix}(ax)b^{n-1}+ \cdots +\begin{pmatrix}n \\n-1\end{pmatrix}(ax)^{n-1}b+\begin{pmatrix}n \\n\end{pmatrix}(ax)^{n}\end{align*}$

$\displaystyle \begin{align*}\implies \frac{d(ax+b)^n}{dx}&=0+\begin{pmatrix}n \\1\end{pmatrix}ab^{n-1}+\cdots+\begin{pmatrix}n \\n-1\end{pmatrix}a^{n-1}(n-1)x^{n-2}+\begin{pmatrix}n \\n\end{pmatrix}a^{n}(n)x^{n-1}\end{align*}$

$\displaystyle \begin{align*}\implies \frac{d(ax+b)^n}{dx}&=a\left[0+ \begin{pmatrix}n \\1\end{pmatrix}b^{n-1}+\cdots+\begin{pmatrix}n \\n-1\end{pmatrix}a^{n-2}(n-1)x^{n-2}+\begin{pmatrix}n \\n\end{pmatrix}a^{n-1}(n)x^{n-1} \right]\end{align*}$

$\displaystyle \begin{align*}\implies \frac{d(ax+b)^n}{dx}&=a\cdot n\left[0+ \begin{pmatrix}n-1 \\0\end{pmatrix}b^{n-1}+\cdots+\begin{pmatrix}n-1\\n-2\end{pmatrix}a^{n-2}x^{n-2}+\begin{pmatrix}n-1 \\n-1\end{pmatrix}a^{n-1}x^{n-1} \right]\end{align*}$

$\displaystyle \begin{align*}\implies \frac{d(ax+b)^n}{dx}&=a\cdot n\left[\sum_{i=0}^{n-1}\begin{pmatrix}n-1\\i \end{pmatrix}(ax)^{i}b^{(n-1-i)}\right]\end{align*}$

$\displaystyle \begin{align*}\implies \frac{d(ax+b)^n}{dx}&=a\cdot n(ax+b)^{n-1}\end{align*}$

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

Or you can use induction to prove that the derivative of $\displaystyle (ax+ b)^n$ is $\displaystyle na(ax+ b)^{n-1}$.

If n= 1, the derivative of [tex](ax+ b)^1= ax+ b[tex] is $\displaystyle a= (1)a(ax+ b)^0$.

Assume that, for some k, the derivative of $\displaystyle (ax+ b)^k$ is $\displaystyle ka(ax+ b)^{k-1}$.

Then $\displaystyle (ax+ b)^{k+1}= (ax+ b)^k(ax+ b)$ and, by the product rule, the derivative is $\displaystyle na(ax+ b)^{k-1}(ax+ b)+ (ax+ b)^k(a)= na(ax+ b)^k+ a(ax+ b)^k= (n+1)a(ax+ b)^k$

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

$\displaystyle \frac{d}{dx}x^n=nx^{n-1}$

Using the Chain Rule

$\displaystyle u=ax+b$

$\displaystyle \frac{d}{dx}u^n=\frac{du}{dx}\;\frac{d}{du}u^n=(a) nu^{n-1}$

$\displaystyle =an(ax+b)^{n-1}$

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.

Quote:

To expand $\displaystyle (ax+b)^n$ you will need to use the binomial theorem.

$\displaystyle (ax+b)^n=\sum_{i=0}^{n}\begin{pmatrix}n \\ i\end{pmatrix}(ax)^{i}b^{(n-i)}$

$\displaystyle \begin{align*}\implies (ax+b)^n &=\begin{pmatrix}n \\0\end{pmatrix}b^n+\begin{pmatrix}n \\1\end{pmatrix}(ax)b^{n-1}+ \cdots +\begin{pmatrix}n \\n-1\end{pmatrix}(ax)^{n-1}b+\begin{pmatrix}n \\n\end{pmatrix}(ax)^{n}\end{align*}$

$\displaystyle \begin{align*}\implies \frac{d(ax+b)^n}{dx}&=0+\begin{pmatrix}n \\1\end{pmatrix}ab^{n-1}+\cdots+\begin{pmatrix}n \\n-1\end{pmatrix}a^{n-1}{\color{red}b}(n-1)x^{n-2}+\begin{pmatrix}n \\n\end{pmatrix}a^{n}(n)x^{n-1}\end{align*}$

$\displaystyle \begin{align*}\implies \frac{d(ax+b)^n}{dx}&=a\left[0+ \begin{pmatrix}n \\1\end{pmatrix}b^{n-1}+\cdots+\begin{pmatrix}n \\n-1\end{pmatrix}a^{n-2}{\color{red}b}(n-1)x^{n-2}+\begin{pmatrix}n \\n\end{pmatrix}a^{n-1}(n)x^{n-1} \right]\end{align*}$

$\displaystyle \begin{align*}\implies \frac{d(ax+b)^n}{dx}&=a\cdot n\left[0+ \begin{pmatrix}n-1 \\0\end{pmatrix}b^{n-1}+\cdots+\begin{pmatrix}n-1\\n-2\end{pmatrix}a^{n-2}{\color{red}b}x^{n-2}+\begin{pmatrix}n-1 \\n-1\end{pmatrix}a^{n-1}x^{n-1} \right]\end{align*}$

$\displaystyle \begin{align*}\implies \frac{d(ax+b)^n}{dx}&=a\cdot n\left[\sum_{i=0}^{n-1}\begin{pmatrix}n-1\\i \end{pmatrix}(ax)^{i}b^{(n-1-i)}\right]\end{align*}$

$\displaystyle \begin{align*}\implies \frac{d(ax+b)^n}{dx}&=a\cdot n(ax+b)^{n-1}\end{align*}$