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Math Help - Differentiating Sin Cos

  1. #1
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    Differentiating Sin Cos

    Help!! Just dabbling into differentiation...

    f(x) = cos (3x)

    f'(x) = -sin (3x)

    RIght?
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  2. #2
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    You need to use the chain rule to do this.
    until your good with the rule do it manually using the formula below.

    \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

    so let u = 3x
    then you got y = \cos u

    then use the chain rule to find \frac{dy}{dx}
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  3. #3
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    I think I got this one right?

    k(x) = x^7 cos(4x)

    f(x) = x^2
    f ' (x) = 2x

    g(x) = cos(4x)
    g ' (x) = -4sin(4x)

    by the product rule:


    k'(x) = f '(x) g (x) + f (x) g'(x)
    = (2x) (cos (4x) + x^2 (-4sin(4x))

    I can't simplify that anymore, right?
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  4. #4
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    Quote Originally Posted by Sweeties View Post
    I think I got this one right?

    k(x) = x^7 cos(4x)

    f(x) = x^2
    f ' (x) = 2x

    g(x) = cos(4x)
    g ' (x) = -4sin(4x)

    by the product rule:


    k'(x) = f '(x) g (x) + f (x) g'(x)
    = (2x) (cos (4x) + x^2 (-4sin(4x))

    I can't simplify that anymore, right?
    Be careful with your parenthesis

    if you wrote
     2x( \cos 4x - 4x^2 \sin 4x ) which is what i am reading your wrong.

    but I am fairly sure you meant.

     2x \cos 4x - 4x^2 \sin 4x which is correct, good stuff.
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  5. #5
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    I thik my basic maths skills let me down more than the ability to grasp higher concepts

    I have differentiated a question using quotient rule and got as far as:

    k'(x) = ((4-t^2) (3e^(3t)) - (e^(3t)) (-2t)) / (4-t^2)^2

    I put it into QuickMath but it comes up all funny

    so, what next? can i simplify this further, if so how? thanks.
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  6. #6
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    I have no idea what the question is, but from your answer i guess you have been told to differentiate k(t) = \frac{ e^{3t}}{4-t^2}

    I am finding it difficult to read your answer, (maybe you should try using LaTeX) but it looks like you have written. k'(t) = \frac{ 3e^{3t}(4-t^2) + 2t \cdot e^{3t}}{(4-t^2)^2}
    which is correct.
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  7. #7
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    Quote Originally Posted by bobak View Post
    I have no idea what the question is, but from your answer i guess you have been told to differentiate k(t) = \frac{ e^{3t}}{4-t^2}

    I am finding it difficult to read your answer, (maybe you should try using LaTeX) but it looks like you have written. k'(t) = \frac{ 3e^{3t}(4-t^2) + 2t \cdot e^{3t}}{(4-t^2)^2}
    which is correct.
    Yes, that's the question. Sorry, I should have written it out in teh first place.
    I have Googled latex and will use it in future.

    So the answer can't be simplified further?
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  8. #8
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    well you can take out a factor of e^{3t} the numerator. Can't see anything else.
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  9. #9
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    ..
    Attached Thumbnails Attached Thumbnails Differentiating Sin Cos-gash.png  
    Last edited by CaptainBlack; February 17th 2008 at 12:58 PM.
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  10. #10
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    ...sorry. just getting used to all this!
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  11. #11
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    I hope you didn't pay for QuickMath, because it is wrong.

    ln( 3 \sin (6x) ) = \ln (3) + \ln ( \sin (6x) )

    \frac{d}{dx} \left ( \ln (3) + \ln ( \sin (6x) ) \right ) = 0 + \frac{6 \cos (6x) }{ \sin (6x)} = 6 \cot (6x)
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  12. #12
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    Quote Originally Posted by bobak View Post
    I hope you didn't pay for QuickMath, because it is wrong.

    ln( 3 \sin (6x) ) = \ln (3) + \ln ( \sin (6x) )

    \frac{d}{dx} \left ( \ln (3) + \ln ( \sin (6x) ) \right ) = 0 + \frac{6 \cos (6x) }{ \sin (6x)} = 6 \cot (6x)
    OK. Can you explain how you go did this step?

    ln( 3 \sin (6x) ) = \ln (3) + \ln ( \sin (6x) )

    -Don't forget, I'm new to all this! Be gentle please!
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  13. #13
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    you use the laws if logs. Are you fimmilar with them ? ( you should have been taught them in your C2 math module).

    \log ab = \log a + \log b
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  14. #14
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    Quote Originally Posted by bobak View Post
    you use the laws if logs. Are you fimmilar with them ? ( you should have been taught them in your C2 math module).

    \log ab = \log a + \log b
    Oh yes! I forgot. Thank you.

    Also, you have used the chain rule in your solution and I need to use the composite rule:
    k'(x)=g'(f(x))f'(x)

    which I think is more complicated, and I can't seem to get the same answer as you got. I will have another go now that you've reminded me of the log rule
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  15. #15
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    Quote Originally Posted by Sweeties View Post
    Oh yes! I forgot. Thank you.

    Also, you have used the chain rule in your solution and I need to use the composite rule:
    k'(x)=g'(f(x))f'(x)

    which I think is more complicated, and I can't seem to get the same answer as you got. I will have another go now that you've reminded me of the log rule
    I am guessing that your reading the rules out of your C3 book.

    One think that you need to realise that all the rules different repertations of the chain rules are infact the same.

    I advise you use the chain rule as \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}<br />
and with you will be able to do this quickly and accurately.

    so in this case we have y = \ln ( \sin (6x) )

    u = \sin (6x)
    \frac{du}{dx} = 6 \cos (6x)
    y = \ln u
    find dy/du

    then finish it off
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