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Math Help - Derive this?

  1. #1
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    Derive this?

    f(x)=\displaystyle\sqrt{49-x^2}+7 \cos^{-1}\left(\frac{x}{7}\right)

    So that F'(x) has this format:

    \displaystyle -\left(\frac{x+c}{mx+n}\right)^{p}

    When I derive this, I get:
    \diplaystyle\frac{-x}{\sqrt{(49-x^2})}+\frac{1}{\sqrt{(1-(\frac{x}{7})^2)}}

    How do I get in into that format?
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  2. #2
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    Quote Originally Posted by Vamz View Post
    f(x)=\displaystyle\sqrt{49-x^2}+7 \cos^{-1}\left(\frac{x}{7}\right)

    So that F'(x) has this format:

    \displaystyle -\left(\frac{x+c}{mx+n}\right)^{p}

    When I derive this, I get:
    \diplaystyle\frac{-x}{\sqrt{(49-x^2})}+\frac{1}{\sqrt{(1-(\frac{x}{7})^2)}}

    How do I get in into that format?

    \diplaystyle \frac{-x}{\sqrt{49-x^2}}- \frac{7}{7} \cdot \frac{1}{\sqrt{1-(\frac{x}{7})^2}}

    \diplaystyle \frac{-x}{\sqrt{49-x^2}}- \frac{7}{\sqrt{49} \cdot \sqrt{1-(\frac{x}{7})^2}}

    \diplaystyle \frac{-x}{\sqrt{49-x^2}}- \frac{7}{\sqrt{49-x^2}}

    \diplaystyle \frac{-(x+7)}{\sqrt{49-x^2}}
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  3. #3
    Member SENTINEL4's Avatar
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    You get the derivative -\frac{x}{\sqrt{49-x^2}}-\frac{1}{\sqrt{1-\frac{x^2}{49}}} = -\frac{x}{\sqrt{49-x^2}}-\frac{1}{\sqrt{\frac{49-x^2}{49}}} = -\frac{x}{\sqrt{49-x^2}}-\frac{1}{\frac{\sqrt{49-x^2}}{7}}

    From there by doing some simple things you can get the form you want
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  4. #4
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    Thanks!
    I don't think that is quite right however!

    There needs to be a "p" value & it cannot be "1"
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  5. #5
    Member SENTINEL4's Avatar
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    If you do some things more you will get a value for p that isn't 1
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  6. #6
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    Please elaborate?
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  7. #7
    Member SENTINEL4's Avatar
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    -\frac{x}{\sqrt{49-x^2}}-\frac{7}{\sqrt{49-x^2}} = -\frac{x+7}{\sqrt{49-x^2}} = -\frac{x+7}{\sqrt{7-x}\sqrt{7+x}}

    one-two more steps and you'll get what you want
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  8. #8
    Super Member Aryth's Avatar
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    Quote Originally Posted by Vamz View Post
    f(x)=\displaystyle\sqrt{49-x^2}+7 \cos^{-1}\left(\frac{x}{7}\right)

    So that F'(x) has this format:

    \displaystyle -\left(\frac{x+c}{mx+n}\right)^{p}

    When I derive this, I get:
    \diplaystyle\frac{-x}{\sqrt{(49-x^2})}+\frac{1}{\sqrt{(1-(\frac{x}{7})^2)}}

    How do I get in into that format?
    You're derivative is equivalent to what they have except at one point. The derivative of the inverse cosine has a negative on it, not a positive, turning your answer to:

    \frac{-x}{\sqrt{49 - x^2}} + \frac{-1}{\sqrt{1 - \left(\frac{x}{7}\right)^2}}

    Then you take the second term and multiply it by 1 = \frac{7}{7} to get:

    -\frac{x}{\sqrt{49-x^2}}-\frac{7}{\sqrt{49-x^2}}

    Then you put everything together to get:

    -\frac{x+7}{\sqrt{49-x^2}}

    We know that the thing inside the square root can factor:

    -\frac{x + 7}{\sqrt{7-x}\sqrt{7+x}}

    From there, you simply recognize that:

    \frac{x}{\sqrt{x}} = \sqrt{x}

    To get:

    -\frac{\sqrt{7 + x}}{\sqrt{7 - x}} = -\sqrt{\frac{7+x}{7-x}}

    This will then give:

    -\left(\frac{7+x}{7-x}\right)^{\frac{1}{2}}

    So, the values are:

    c = 7
    m = -1
    n = 7
    p = 1/2
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