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Math Help - Did I derive this right?

  1. #1
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    Did I derive this right?



    Where did I go wrong?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by kmjt View Post


    Where did I go wrong?
    For a start:

    How do you get from the first line to the second?

    CB
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    For somee reason I thought you can bring the denominator up and attach it, I guess you can't
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    Quote Originally Posted by kmjt View Post
    For somee reason I thought you can bring the denominator up and attach it, I guess you can't
    well you can if you raise the denominators to the power of -1

    i.e.  A = x^{2}.4\pi^{-1} + (x^{2} - 80x + 1600).16^{-1} but thats not reall that useful for what you want to do.

    Just use the quotient rule twice and sum the two answes to get  A'

    Cant see how this is university level though...
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  5. #5
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by kmjt View Post


    Where did I go wrong?
    Notice that 4\pi \text{and} 16 in the denominators of the first and second expressions are constants. So your derivative would be:

    A' = \frac{1}{4 \pi} \frac{d}{dx} x^2 + \frac{1}{16} \frac{d}{dx} (x^2-80x+1600)
    Last edited by harish21; May 1st 2010 at 10:37 AM.
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  6. #6
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    So far I used the quotient rule on x^2/4pi and got 8pi x / 4pi^2, is that correct?
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    Quote Originally Posted by kmjt View Post
    So far I used the quotient rule on x^2/4pi and got 8pi x / 4pi^2, is that correct?

    Nope, try again. i'll tell you the answer after another effort. Using Harish's method is easier perhaps
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    MHF Contributor harish21's Avatar
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    Quote Originally Posted by kmjt View Post
    So far I used the quotient rule on x^2/4pi and got 8pi x / 4pi^2, is that correct?
    No. Look at the post above. Because 4pi and 16 are constants, you can isolate those terms while differentiating

    your answer would be:

    [(1/4pi) times the derivative of x^2 ] +

    [(1/16) times the derivative of x^2-80x+1600]
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    I think I should stick to differentiating the left and than the right, because that's what we do in class. Heres where I got my answer from doing the quotient rule on x^2 / 4pi:

    f'(x) = g'(x)h(x) - g(x)h'(x) / (4pi)^2

    = (2x)(4pi) - (x^2)(0) / 4pi^2

    = 8pix - 0 / 4pi^2

    = 8pix / 4pi^2

    Where did I mess up? I used a derivative calculator online and it said the derivative would be x / 2pi

    Sorry again harish thanks for that technique but I don't believe I should be doing it that way in the class im in
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  10. #10
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    Quote Originally Posted by kmjt View Post
    I think I should stick to differentiating the left and than the right, because that's what we do in class. Heres where I got my answer from doing the quotient rule on x^2 / 4pi:

    f'(x) = g'(x)h(x) - g(x)h'(x) / (4pi)^2

    = (2x)(4pi) - (x^2)(0) / 4pi^2

    = 8pix - 0 / 4pi^2

    = 8pix / 4pi^2

    Where did I mess up? I used a derivative calculator online and it said the derivative would be x / 2pi

    You have to remember that  (4\pi)^{2} = 4\pi.4\pi = 16\pi^{2} and NOT equal to  4\pi^{2}

    Sorry again harish thanks for that technique but I don't believe I should be doing it that way in the class im in
    Ok here goes...

    let  u = x^{2} and  v = 4\pi

    differentiating u w.r.t x we get  \frac{du}{dx} = 2x and
    differentiating v w.r.t x we get  \frac{dv}{dx} = 0 This is always the case when differentiating constants.

    Applying the quotient rule we get;

     \frac{dy}{dx} = \frac{4\pi.2x - x^{2}.0}{4\pi^{2}}

    =>  \frac{dy}{dx} = \frac{8\pi.x}{(4\pi)^{2}}

    =>  \frac{dy}{dx} = \frac{8\pi.x}{16\pi^{2}}

    Now we can cancel out  8\pi as it is common to both the top and bottom line.

    This leaves us with  \frac{dy}{dx} = \frac{x}{2\pi}
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  11. #11
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    Quote Originally Posted by kmjt View Post
    I think I should stick to differentiating the left and than the right, because that's what we do in class. Heres where I got my answer from doing the quotient rule on x^2 / 4pi:

    f'(x) = g'(x)h(x) - g(x)h'(x) / (4pi)^2

    = (2x)(4pi) - (x^2)(0) / 4pi^2 ... denominator should be \textcolor{red}{16 \pi^2}<br />
    ...
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  12. #12
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    Quote Originally Posted by kmjt View Post
    I think I should stick to differentiating the left and than the right, because that's what we do in class. Heres where I got my answer from doing the quotient rule on x^2 / 4pi:

    f'(x) = g'(x)h(x) - g(x)h'(x) / (4pi)^2

    = (2x)(4pi) - (x^2)(0) / 4pi^2

    = 8pix - 0 / 4pi^2

    = 8pix / 4pi^2

    Where did I mess up? I used a derivative calculator online and it said the derivative would be x / 2pi

    Sorry again harish thanks for that technique but I don't believe I should be doing it that way in the class im in
    \frac{1}{4\pi} \frac{d}{dx}x^2 = \frac{1}{4 \pi} 2x = \frac{2x}{4 \pi} = \frac{x}{2 \pi}

    that is the derivative of the first expression of your term(note that this is the same as what you got from the derivative calculator).. you have to do the same to find the derivative of the second term, and add those two to get your answer..
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  13. #13
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    I see thanks
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