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Math Help - [SOLVED] Derivative of an inverse trigonometric function

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    [SOLVED] Derivative of an inverse trigonometric function

    I need help deriving this function:

    g(t)=tan^-1 (t/ sqrt(2 - t^2))

    Any help would be appreciated! Thanks


    What I got when I tried deriving it:
    1/(1+(t/sqrt(2-t^2))^2) x (sqrt(2-t^2)+(t^2/(sqrt(2-t^2)))/(sqrt(2-t^2))
    and eventually ended up with:
    2/((sqrt(2-t^2))^3 + t^3(sqrt(2-t^2)))
    Last edited by yzobel; November 19th 2009 at 05:23 PM.
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  2. #2
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    Quote Originally Posted by yzobel View Post
    I need help deriving this function:

    g(t)=tan^-1 (t/ sqrt(2 - t^2))

    Any help would be appreciated! Thanks
    \frac{d}{dt} \tan^{-1}(u) = \frac{1}{1+u^2} \cdot \frac{du}{dt}
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  3. #3
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    Quote Originally Posted by skeeter View Post
    \frac{d}{dt} \tan^{-1}(u) = \frac{1}{1+u^2} \cdot \frac{du}{dt}
    Thank you, but I already tried that before and came up with a long, complicated answer. My teacher mentioned that g should be simplified beforehand but I'm not quite sure how to do that.
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    Isn't {tan^-1(x)} just cot(x)?
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  5. #5
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    Quote Originally Posted by RockHard View Post
    Isn't {tan^-1(x)} just cot(x)?
    No, \tan^{-1} x = \arctan x
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    Yes, I know that... tan^-1(x) can be also written as \frac{1}{tan(x)}
    which by trigonmetric identities cot(x) = \frac{1}{tan(x)}, I just did not want to state something and be wrong.

    Table of Trigonometric Identities

    Also it is whatever choice maybe the simplest because \int cot(x) can be written as \int\frac{cos(x)}{sin(x)} and from your problem above

    \int\frac{cos(x)}{sin(x)\sqrt{2-t^2}}

    And from here you can just use a trig. substitution using the identity, that fits the description of what's under the square root.
    sin^2(x)+cos^2(x) = 1
    Last edited by RockHard; November 19th 2009 at 06:18 PM.
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  7. #7
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    Quote Originally Posted by RockHard View Post
    Yes, I know that... tan^-1(x) can be also written as \frac{1}{tan(x)}
    which by trigonmetric identities cot(x) = \frac{1}{tan(x)}, I just did not want to state something and be wrong.

    Table of Trigonometric Identities
    They're not the same.

    tan^-1(x)=arctanx

    (tanx)^-1 =cotx
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  8. #8
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    Ah, simply overlooked, it my mistake good sir.
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  9. #9
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    Quote Originally Posted by yzobel View Post
    Thank you, but I already tried that before and came up with a long, complicated answer. My teacher mentioned that g should be simplified beforehand but I'm not quite sure how to do that.
    The answer is not long and is not complicated.

    After a small amount of algebra (at this level it should be seen as small):

    \frac{1}{1 + u^2} = \frac{2 - t^2}{2}

    \frac{du}{dt} = \frac{2}{(2 - t^2) \sqrt{2 - t^2}}.

    And thus the final answer for \frac{dg}{dt} is simple.
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