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Math Help - Two derivative problems

  1. #1
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    Two derivative problems

    So, my homework problem is deriving -2*sin(x)*cos(x) as part of using the concavity theorem on -6x^2+cos^2(x). My calculator is saying 2-4(cos(x))^2, but I'm not getting to the same answer.


    My other question is working the problem sin^2(cos(4x)), this was a problem on my exam that I got right, but only because I memorized the general pattern of the problem when we got it in class the week before and she just happened to have it on the test. I need to see it worked out step by step, as I get lost in it pretty quickly.

    ( It works out to -8*sin(4x)*sin(cos(4x)) * cos(cos(4x)) )

    Thanks!
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Wolvenmoon View Post
    So, my homework problem is deriving -2*sin(x)*cos(x) as part of using the concavity theorem on -6x^2+cos^2(x). My calculator is saying 2-4(cos(x))^2, but I'm not getting to the same answer.


    My other question is working the problem sin^2(cos(4x)), this was a problem on my exam that I got right, but only because I memorized the general pattern of the problem when we got it in class the week before and she just happened to have it on the test. I need to see it worked out step by step, as I get lost in it pretty quickly.

    ( It works out to -8*sin(4x)*sin(cos(4x)) * cos(cos(4x)) )

    Thanks!
    Use an identity:

    -2\sin(x)\cos(x) = -\sin(2x)

    -\frac{d}{dx} \sin(2x) = -2\cos(2x)

    \cos(2x) = 2\cos^2(x)-1 so -2\cos(2x) = -4\cos^2(x)+2 = 2-4\cos^2(x) which is the calculator's reply.


    =================================

    You can also use the product rule

    -2 \frac{d}{dx} = \sin(x)\cos(x)

    u = \sin(x) \: \: \rightarrow \: \: u' = \cos(x)

    v = \cos(x) \: \: \rightarrow \: \: v' = -\sin(x)

    By the product rule

    -2(-\sin^2(x)+\cos^2(x)) = 2\sin^2(x)-2\cos^2(x)

    However, given that \sin^2(x) = 1-\cos^2(x)

    2-2\cos^2(x)-2\cos^2(x) = 2-4\cos^2(x)
    Last edited by e^(i*pi); March 15th 2010 at 12:55 PM. Reason: Latex
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  3. #3
    Member mathemagister's Avatar
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    Quote Originally Posted by Wolvenmoon View Post
    So, my homework problem is deriving -2*sin(x)*cos(x) as part of using the concavity theorem on -6x^2+cos^2(x). My calculator is saying 2-4(cos(x))^2, but I'm not getting to the same answer.


    My other question is working the problem sin^2(cos(4x)), this was a problem on my exam that I got right, but only because I memorized the general pattern of the problem when we got it in class the week before and she just happened to have it on the test. I need to see it worked out step by step, as I get lost in it pretty quickly.

    ( It works out to -8*sin(4x)*sin(cos(4x)) * cos(cos(4x)) )

    Thanks!
    For your 2nd question (which hasn't been answered):

    Use:

    The Chain Rule

    \boxed{\frac{d}{dx}f\left[g(x)\right] = f'\left[g(x)\right]\cdot g'(x)}

    Quote Originally Posted by Wolvenmoon View Post
    I memorized the general pattern of the problem
    If this is the "general pattern" you memorized, you did the right thing.

    In this case you want to differentiate  \sin^2[\cos(4x)]

    Using the chain rule: f(x) = \sin^2x, \ \ g(x) = \cos(4x)

    Note that f(x) is itself a combo function with the \sin{x} trapped inside the (\sin{x})^2

    So the derivative of (\sin{x})^2 is f'(x) = 2\sin{x} \cdot \frac{d}{dx} \sin{x} = 2\sin{x}\cos{x}

    A trig identity says that 2\sin{x}\cos{x} = \sin{(2x)}

    So f'(x) = \sin{(2x)}

    Notice the g(x) is also a combo with the 4x trapped inside the \cos{(4x)}.

    So the derivative of \cos{(4x)} is g'(x) = -\sin{(4x)} \cdot 4 = -4\sin{(4x)}

    Now using the master chain rule:

    The derivative of sin^2[cos(4x)] is f'[g(x)]\cdot g'(x)

    Substitute g'(x) and substitute g(x) for x in the f'(x) calculation:

    \sin(2cos(4x))\cdot -4sin(4x) = -4\sin[2\cos{(4x)}] \sin{(4x)}

    Phew! And that's the answer.

    NOTE: This answer is a bit more simplified that the one you got because I used the trig identity 2\sin{x}\cos{x} = \sin{(2x)}. You can check that they are both the same function, by graphing them or randomly checking points such as x=3. I have checked and saw that they are the same, so my answer is also correct.

    Hope that helps

    Mathemagister
    Last edited by mathemagister; March 31st 2010 at 04:39 AM.
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