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Math Help - Derivative of inverse sine?

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    Derivative of inverse sine?

    I have been getting mixed signals about what the derivative of arcsin(u) is from my math teacher. He says that \frac{d}{du}[arcsin(u)] \: = \: \frac{du}{\sqrt{1 - u^{2}}}, and that \int \frac{du}{\sqrt{a^{2} + u^{2}}} \: = \: arcsin(\frac{u}{a}) , where a is a constant.
    I realized these are not the same derivative of arcsin(u), so I checked Wolfram Alpha and it said that the integral version that I have been given actually solves for the inverse hyperbolic sine of u. Which is right?
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    Re: Derivative of inverse sine?

    Quote Originally Posted by j1234 View Post
    I have been getting mixed signals about what the derivative of arcsin(u) is from my math teacher. He says that \frac{d}{du}[arcsin(u)] \: = \: \frac{du}{\sqrt{1 - u^{2}}}, and that \int \frac{du}{\sqrt{a^{2} + u^{2}}} \: = \: arcsin(\frac{u}{a}) , where a is a constantt?
    Look at this \frac{d}{{du}}\left[ {\arcsin \left( {\frac{u}{a}} \right)} \right] = \frac{{\frac{1}{a}}}{{\sqrt {1 - \frac{{{u^2}}}{{{a^2}}}} }} = \frac{1}{{\sqrt {{a^2} - {u^2}} }}

    It is clear that a simple sign error was made in your instructor's example.
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