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Math Help - Trigonometric Derivative

  1. #1
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    Trigonometric Derivative

    Hello,

    I am doing a list of problems to do and have experienced some difficulty doing a particular one.

    It asks to find the derivative of
    f(x) = sin^2 (1+cos^3 (1+tan^4 (1+x))).

    I've added a space ahead of exponents to avoid confusion, but it is all together.

    Now, I don't know whether to use the chain rule multiple times, or how that would work, or anything. I don't know whether to see "(1+cos^3 (1+tan^4 (1+x)))" to sin^2 as you would see it as sin^2 x or something. I'm very confused.

    Any help would be greatly appreciated, thank you!
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  2. #2
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    By exponents do you mean brackets? If so then it might help to know you can re-write it as

    f(x) = [ sin( 1 + {cos( 1 + [tan(1+x)]^4) }^3) ]^2

    and yes, the chain rule nested

    Here's a few more of the pieces...

    \frac{d}{dx}\ \tan(1+x) = \sec^2(1+x)

    ... and just in case a picture helps...



    ... so...

    \frac{d}{dx}\ (1 + \tan^4(1+x)) = 4 \tan^3(1+x) * \sec^2(1+x)



    ... so...

    \frac{d}{dx}\ \cos(1 + \tan^4(1+x))

     = -\sin(1 + \tan^4(1+x))\ 4 \tan^3(1+x)\ \sec^2(1+x)



    ... so...

    \frac{d}{dx}\ [1 + \cos^3(1 + \tan^4(1+x))]

     = 3 \cos^2(1 + \tan^4(1+x))(-\sin(1 + \tan^4(1+x)))\ 4 \tan^3(1+x)\ \sec^2(1+x)

    Etc...

    Nesty nasty! Excuse myriad errors, hopefully fixed.

    Or perhaps you meant exponents after all... loads of fun then!

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    Last edited by tom@ballooncalculus; November 26th 2009 at 01:51 AM.
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  3. #3
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    Hello,

    Your answer is much appreciated. Thank you very much.

    Just one more simple question, now that I have the derivative
    \frac{d}{dx}\ [1 + \cos^3(1 + \tan^4(1+x))]

    should I continue to work at my answer such that I use the product rule to find

    \frac{d}{dx}\ sin^2 * [1 + \cos^3(1 + \tan^4(1+x))] + \frac{d}{dx}\ [1 + \cos^3(1 + \tan^4(1+x))] *  sin^2 ?
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  4. #4
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    Quote Originally Posted by Torcida1911 View Post
    Just one more simple question,
    Unless I'm missing a trick*, there's no way to keep it simple! This problem is unusually messy, but I guess the idea is that by the end you know the chain rule...

    Not product rule - remember sin^2(x) means (sin(x))^2

    There are 2 more layers to deal with: sin and then squaring.

    \frac{d}{dx}\ \sin[1 + \cos^3(1 + \tan^4(1+x))]

    will be cos of the same square-bracket-ed expression and then that multiplied by the derivative of same square-bracket-ed expression - which is the last result we got.

    For what it's worth, I'll do a pic working from the outside in...

    *We could maybe use some identities before starting to differentiate... not sure... don't think so

    OK, here's the big pic, in a 'spoiler' - on this occasion so we don't spoil the page layout rather than the fun. The whole derivative should be the product of all the left forks. You can simplify that result, a bit.

    Spoiler:
    Last edited by tom@ballooncalculus; November 26th 2009 at 09:04 AM.
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  5. #5
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    I've got it.. what an intense problem..

    Thank you so much for your time and effort. Great visuals.
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