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Math Help - Taking The Derivative

  1. #1
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    Taking The Derivative

    What is the derivative of this equation:
    y= arcsin (radical(2t))
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  2. #2
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    The chain rule is

     \frac{dy}{dt} = \frac{dy}{du}\times \frac{du}{dt}

    For your function y = \arcsin(\sqrt{2t}) make u = \sqrt{2t} so y = \arcsin(u)

    Now find  \frac{dy}{du} and  \frac{du}{dt}
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  3. #3
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    hi
    just to add, \forall x\in (-1,1),\arcsin'(x)=\frac{1}{\sqrt{1-x^2}}
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    Then what is the derivative of (2t^.5)
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  5. #5
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    When i took the derivative of (2t^.5) i got 1/(2t^.5)
    does that even make sense?
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  6. #6
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    Quote Originally Posted by emilyroy10 View Post
    Then what is the derivative of (2t^.5)
    \frac{d}{dt}(2 \sqrt{t}) = \frac{1}{\sqrt{t}}
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  7. #7
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    what is the derivative of ((2t)^.5)
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  8. #8
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    (\sqrt{2t})'=\frac{1}{2}(2t)^{-\frac{1}{2}}\times2=\frac{1}{\sqrt{2t}}
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  9. #9
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    okay thats what i got when i took the derivative of ((2t)^.5), which was 1/((2t)^.5).
    but what do you get when you multiply it by the derivative of arcsin, which is 1/((1-((2t)^2))^.5)
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  10. #10
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    you should get, \frac{1}{\sqrt{2}\sqrt{1-2t}\sqrt{t}}
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