# Thread: proof: derivative for (ax + b)^n

1. ## 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

2. ## 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*}

3. ## 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 $\displaystyle na(ax+ b)^{k-1}(ax+ b)+ (ax+ b)^k(a)= na(ax+ b)^k+ a(ax+ b)^k= (n+1)a(ax+ b)^k$

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

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

Using the Chain Rule

$\displaystyle u=ax+b$

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

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

5. ## 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?

6. You could also try

$\displaystyle \frac{d}{dx}\left[ax+b\right]^n=\frac{d}{dx}\left[\binom{n}{0} b^n+\binom{n}{1} (ax)b^{n-1}+\binom{n}{2} (ax)^2b^{n-2}+....+\binom{n}{n}(ax)^n\right]$

$\displaystyle =(na)b^{n-1}+\binom{n}{2}2a^2xb^{n-2}+\binom{n}{3}3a^3x^2b^{n-3}+...\binom{n}{n}na^nx^{n-1}$

Now using $\displaystyle \binom{n}{p}=\frac{n}{p}\binom{n-1}{p-1}$ we get the derivative to be

$\displaystyle (na)b^{n-1}+(na)\binom{n-1}{1}(ax)b^{n-2}+(na)\binom{n-1}{2}(ax)^2b^{n-3}+.....+(na)\binom{n-1}{n-1}(ax)^{n-1}$

$\displaystyle =na\left[\binom{n-1}{0}b^{n-1}+\binom{n-1}{1}(ax)b^{(n-1)-1}+\binom{n-1}{2}(ax)^2b^{(n-1)-2}+....+\binom{n-1}{n-1}(ax)^{n-1}\right]$

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

7. ## 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

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

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.

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*}

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