Results 1 to 8 of 8

Math Help - proof: derivative for (ax + b)^n

  1. #1
    Senior Member furor celtica's Avatar
    Joined
    May 2009
    Posts
    271

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member sbhatnagar's Avatar
    Joined
    Sep 2011
    From
    New Delhi, India
    Posts
    200
    Thanks
    17

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

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

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

    \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*}

    \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*}

    \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*}


    \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*}

    \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*}

    \begin{align*}\implies \frac{d(ax+b)^n}{dx}&=a\cdot n(ax+b)^{n-1}\end{align*}
    Last edited by sbhatnagar; December 6th 2011 at 02:54 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,572
    Thanks
    1412

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

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

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

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

    Then (ax+ b)^{k+1}= (ax+ b)^k(ax+ b) and, by the product rule, the derivative is 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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1

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

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

    Using the Chain Rule

    u=ax+b

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

    =an(ax+b)^{n-1}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member furor celtica's Avatar
    Joined
    May 2009
    Posts
    271

    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?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    You could also try

    \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]

    =(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 \binom{n}{p}=\frac{n}{p}\binom{n-1}{p-1} we get the derivative to be

    (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}

    =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]

    =(na)(ax+b)^{n-1}
    Last edited by Ackbeet; December 8th 2011 at 06:51 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member furor celtica's Avatar
    Joined
    May 2009
    Posts
    271

    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
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member sbhatnagar's Avatar
    Joined
    Sep 2011
    From
    New Delhi, India
    Posts
    200
    Thanks
    17

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

    Quote Originally Posted by furor celtica View Post
    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 (ax+b)^n you will need to use the binomial theorem.

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

    \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*}

    \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*}

    \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*}


    \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*}

    \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*}

    \begin{align*}\implies \frac{d(ax+b)^n}{dx}&=a\cdot n(ax+b)^{n-1}\end{align*}
    Last edited by sbhatnagar; December 12th 2011 at 12:48 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivative Proof
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 3rd 2009, 08:49 AM
  2. proof of derivative of a^x
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 6th 2009, 09:47 PM
  3. proof to first derivative
    Posted in the Calculus Forum
    Replies: 0
    Last Post: December 7th 2008, 09:08 PM
  4. Proof of derivative
    Posted in the Calculus Forum
    Replies: 10
    Last Post: May 3rd 2008, 08:24 AM
  5. Derivative proof
    Posted in the Calculus Forum
    Replies: 5
    Last Post: November 18th 2007, 05:34 PM

Search Tags


/mathhelpforum @mathhelpforum