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Math Help - Inverse Trigonmetric functions

  1. #1
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    Inverse Trigonmetric functions

    Find the dy/dx.

    arcsin ( \sqrt{x})
    I got \frac{\sqrt{x}}{\sqrt{1-2t}}
    This, is not correct,

    I used the typical trig formula for this. 1/(1-u^2) * du/dx.

    Thanks!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Truthbetold View Post
    Find the dy/dx.

    arcsin ( \sqrt{x})
    I got \frac{\sqrt{x}}{\sqrt{1-2t}}
    This, is not correct,

    I used the typical trig formula for this. 1/(1-u^2) * du/dx.

    Thanks!
    now how in the world did you end up with t in your answer?

    \frac d{dx} \arcsin u = \frac {u'}{\sqrt{1 - u^2}} where u is a function of x

    now, if u = \sqrt{x}, what is u'?
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    Quote Originally Posted by Jhevon View Post
    now how in the world did you end up with t in your answer?

    \frac d{dx} \arcsin u = \frac {u'}{\sqrt{1 - u^2}} where u is a function of x

    now, if u = \sqrt{x}, what is u'?
    I don't know how I got t in my answer. I made a typo.

    Derivative of u'= \frac{1}{2\sqrt{x}}
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Truthbetold View Post
    I don't know how I got t in my answer. I made a typo.

    Derivative of u'= \frac{1}{2\sqrt{x}}
    ok, so now try again. what's the derivative of the original function?
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    \frac{1}{2\sqrt{x} * \sqrt{1-x}}.. my answer
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Truthbetold View Post
    \frac{1}{2\sqrt{x} * \sqrt{1-x}}.. my answer
    that is correct. you could simplify it, but that's fine
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