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Math Help - derivative of arcsine

  1. #1
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    derivative of arcsine

    I dont understand why when the derivative of arcsine is taken it is
    . Can someone help me understand please!

    Thank you!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by gabet16941 View Post
    I dont understand why when the derivative of arcsine is taken it is
    . Can someone help me understand please!

    Thank you!
    Note that \sin^{-1}x is differentiable on the interval -1<x<1 and x=\sin y is differentiable on the interval -\frac{\pi}{2}<y<\frac{\pi}{2}

    Let y=\sin^{-1}x\implies x=\sin y.

    Now, implicitly differentiating both sides, we have

    1=\cos y\frac{\,dy}{\,dx}\implies\frac{\,dy}{\,dx}=\frac{  1}{\cos y}

    We can divide through by \cos y because \cos y\neq0 on the interval -\frac{\pi}{2}<y<\frac{\pi}{2}. Also \cos y >0 on the interval -\frac{\pi}{2}<y<\frac{\pi}{2}. Now, we can use the identity \cos^2y+\sin^2 y =1 to find \cos y:

    \cos^2y+\sin^2y=1\implies \cos^2y=1-\sin^2y\implies \cos y=\sqrt{1-\sin^2y}

    Thus, we see that \frac{\,dy}{\,dx}=\frac{1}{\sqrt{1-\sin^2y}}

    Since we let x=\sin y, it now implies that \color{red}\boxed{\frac{\,d}{\,dx}\left(\sin^{-1}x\right)=\frac{1}{\sqrt{1-x^2}}}

    Does this make sense?
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  3. #3
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    Yes thank you very much!
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