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Math Help - Trouble with the derivative of inverse trigonometric functions.

  1. #1
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    Trouble with the derivative of inverse trigonometric functions.

    I'm having a hard time with the following problem:

    y=arctan(arcsin(2x))

    Am I supposed to do chain rule?
    Someone please help..
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  2. #2
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    Quote Originally Posted by MadHotThunder View Post
    I'm having a hard time with the following problem:

    y=arctan(arcsin(2x))

    Am I supposed to do chain rule?
    Someone please help..
    Yes. So y' = arctan'[(arcsin(2x))] * arcsin'[(2x)] * 2
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  3. #3
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    Quote Originally Posted by Jameson View Post
    Yes. So y' = arctan'[(arcsin(2x))] * arcsin'[(2x)] * 2
    Thank you so much.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by MadHotThunder View Post
    I'm having a hard time with the following problem:

    y=arctan(arcsin(2x))

    Am I supposed to do chain rule?
    Someone please help..
    \frac{d}{dx}\bigg[\arctan(x)\bigg]=\frac{1}{1+x^2}

    \frac{d}{dx}\bigg[\arcsin(2x)\bigg]=\frac{2}{\sqrt{1-4x^2}}

    So now use the chain rule like you alluded to.
    Last edited by Mathstud28; November 3rd 2008 at 06:19 PM. Reason: typo
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  5. #5
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    Yesssss, I used the actual derivatives when I solved the problem, not just arctan'.

    To the both of you, thank you so much.
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