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Math Help - differentiating arcsinx

  1. #1
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    differentiating arcsinx

    hi, just have a few questions

    use chain rule to find d/dx (1/arcsin(x^2))
    i know d/dx (arcsinx)= 1/(1-x^2)^1/2

    also for x in domain [-1,1] prove that arcsinx + arccosx = pi/2
    for this i took the derivatives of the LHS whiched equaled zero but dont know where to go from here

    thanks
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by b0mb3rz View Post
    hi, just have a few questions

    use chain rule to find d/dx (1/arcsin(x^2))
    i know d/dx (arcsinx)= 1/(1-x^2)^1/2
    Note that if \frac{\,d}{\,dx}\left[\arcsin\!\left(u\right)\right]=\frac{1}{\sqrt{1-u^2}}\cdot\frac{\,du}{\,dx}, then it follows that \frac{\,d}{\,dx}\left[\frac{1}{\arcsin\!\left(u\right)}\right]=\sqrt{1-u^2}\cdot\frac{\,dx}{\,du}

    Can you try this problem now?

    also for x in domain [-1,1] prove that arcsinx + arccosx = pi/2
    for this i took the derivatives of the LHS whiched equaled zero but dont know where to go from here

    thanks
    If you differntiate both sides you get \frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}=0\implies 0=0, which would confirm the identity.

    Does this make sense?
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  3. #3
    Senior Member Pinkk's Avatar
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    You could really substitute \frac{\pi}{2} with any constant and that differentiation would hold true. I'll try to think of a way to show this algebraically, but if you think about it geometrically, it has to equal \frac{\pi}{2} since those two operations are going to give you two angles of a right triangle which will add up to 90 degrees, or \frac{\pi}{2}.
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  4. #4
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    thanks, im still a bit unsure about the first question

    but for the second question that's what i got that it was equal to zero but i wasnt certain that was the answer
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by b0mb3rz View Post
    thanks, im still a bit unsure about the first question
    If y=\frac{1}{\arcsin\!\left(x^2\right)}, it follows that y^{\prime}=\frac{1}{\frac{1}{\sqrt{1-\left(x^2\right)^2}}\cdot2x}=\boxed{\frac{\sqrt{1-x^4}}{2x}}

    Does this make sense?
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