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Math Help - Cosine Function Derivative

  1. #1
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    Cosine Function Derivative

    Question # 2 lol. I don't know why I'm having such a hard time getting this....

    Determine the derivative of the following fuction:
    g(x) = (cosx)^3

    I started by using the power rule to come up with:

    g '(x) = 3(cosx)^2
    = 3(-sinx)^2

    According to my book this is wrong though... it has:

    g '(x) = (3(cosx)^2)(-sinx)
    = -3cos^2xsinx

    Which makes no sense to me. Thanks again.
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  2. #2
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    y= \cos ^3x

    using the chain rule \frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}

    so make u = \cos x so y = u^3

    now

    \frac{dy}{du}= 3u^2

    and

    \frac{du}{dx}= -\sin x

    \frac{dy}{dx} = 3u^2 \times -\sin x= 3\cos^2x \times -\sin x = -3\cos^2x\sin x
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  3. #3
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    I was taught the chain rule a little differently. It was shown to me like this:

    f (x) = a^g(x)
    f '(x) = (a^g(x))(ln a)(g '(x))

    Would it be possible to quickly go over how the method you are using works:

    dy/dx = (dy/du)(du/dx)

    Thanks!
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  4. #4
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    Quote Originally Posted by Jools View Post

    Would it be possible to quickly go over how the method you are using works:

    dy/dx = (dy/du)(du/dx)

    Thanks!
    This is exactly what I have done above. This is the chain rule.

    Quote Originally Posted by Jools View Post
    I was taught the chain rule a little differently. It was shown to me like this:

    f (x) = a^g(x)
    f '(x) = (a^g(x))(ln a)(g '(x))
    This is just a version of the chain rule, it won't work in your case because your function is not in the correct form to use this adaptation.
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  5. #5
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    Sorry, I'm confused. Let's take it from the top. First off how did you come up with y = cos^3 x? Then I think I just need to know what each variable in the formula represents and I should be able to figure this out. I appreciate your patience.
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  6. #6
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    Quote Originally Posted by Jools View Post
    I was taught the chain rule a little differently. It was shown to me like this:

    f (x) = a^g(x)
    f '(x) = (a^g(x))(ln a)(g '(x))
    This technique only works if a is a constant.

    I find it easiest to remember this:

    if f(x) =[g(x)]^a, then
    f'(x) = a*[g(x)]^(a-1)*g'(x)

    Here's an examples:

    f(x) = (3x)^3
    f'(x) = 3*(3x)^2 *3 = 81x^2

    Another example:

    f(x) = (x^2)^3
    f'(x) = 3*(x^2)^2*2x = 6x^5

    See how it works? So applying it to your problem:

    f(x) = [cos(x)]^3
    f'(x) = 3* [cos(x)]^2 * (-sin(x))
    = -3 cos^2(x) sin(x)
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  7. #7
    MHF Contributor ebaines's Avatar
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    Quote Originally Posted by Jools View Post
    Sorry, I'm confused. Let's take it from the top. First off how did you come up with y = cos^3 x? Then I think I just need to know what each variable in the formula represents and I should be able to figure this out. I appreciate your patience.
    Ah yes ... the notation cos^3(x) means the the same thing as [cos(x)]^3. Perhaps that's the confusion.
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  8. #8
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    So how exactly do I determine what the g(x) is? For example taking another question asking for the derivative:

    f(x) = sin(x^2)

    Is x^2 the g(x)?
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  9. #9
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    Quote Originally Posted by Jools View Post
    So how exactly do I determine what the g(x) is? For example taking another question asking for the derivative:

    f(x) = sin(x^2)

    Is x^2 the g(x)?
    Yes, that is correct.

    Here is the chain rule written in terms of the composite function:

    \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)
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  10. #10
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    Quote Originally Posted by icemanfan View Post
    Yes, that is correct.

    Here is the chain rule written in terms of the composite function:

    \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)
    Ok I think I'm catching on... So one more...

    g(x) = e^cosx

    How would that one go?
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  11. #11
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    Quote Originally Posted by Jools View Post
    Ok I think I'm catching on... So one more...

    g(x) = e^cosx

    How would that one go?
    Let u = \cos{x} so that g = e^u.


    \frac{du}{dx} = -\sin{x}

    \frac{dg}{du} = e^u = e^{\cos{x}}


    So \frac{dg}{dx} = -\sin{x}\,e^{\cos{x}}.
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  12. #12
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    Quote Originally Posted by Prove It View Post
    Let u = \cos{x} so that g = e^u.


    \frac{du}{dx} = -\sin{x}

    \frac{dg}{du} = e^u = e^{\cos{x}}


    So \frac{dg}{dx} = -\sin{x}\,e^{\cos{x}}.
    Could that be applied to this formula?

    f '(g(x))(g '(x))
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  13. #13
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    Quote Originally Posted by Jools View Post
    Could that be applied to the formula?

    f '(g(x))(g '(x))
    Yes - just replace where I've used u with g(x) and replace where I've used g with f.
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  14. #14
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    Quote Originally Posted by Prove It View Post
    Yes - just replace where I've used u with g(x) and replace where I've used g with f.
    Got it. I REALLY appreciate all the help. I understand how to work out the derivative I think... I'm just having a hard time determining what the value of g(x) is for each variation....
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