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    Hi,

    Given that: y=x-\sqrt {1-x^2} - inverse sin(x)

    show that \frac{dy}{dx} = x (inverse sin(x)) / sqrt(1-x^2)


    I differentiated using product rule but ended up with

    dy/dx = - inverse sin(x) / (2 sqrt(1-x^2))

    (Sorry about the lack of LaTeX, I'm having major issues with it).
    Any help would be appreciated
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    To clarify:

     y = x - \sqrt{1-x^2} - arcsin\ x

    To be shown:

     \frac{dy}{dx} = \frac{x(arcsin\ x)}{\sqrt{1-x^2}}

    right?
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    (Original post by zdo0o)
    To clarify:

     y = x - \sqrt{1-x^2} - arcsin\ x

    To be shown:

     \frac{dy}{dx} = \frac{x(arcsin\ x)}{\sqrt{1-x^2}}

    right?

    Nearly, except in the y = ... it should be sqrt (1-x^2) multiplied by arc sin...
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    Product rule gives:

    dy/dx = 1 - [x(1-x^2)^-0.5 x arcsinx + (1-x^2)^0.5 x 1/sqrt(1-x^2)]

    See if you can go from there.
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    Done, thank you. I forgot about the fact that the bracket was quadratic when I used the product rule.
 
 
 
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Updated: February 1, 2010
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