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Math Help - Inverse of sinx with arcsinx?

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    Inverse of sinx with arcsinx?

    I don't even understand the wording of this. Also arcSINx?

    Denote by y=arcsinx, -1< or = x < or = 1, the inverse of y=sinx with x between -pi/2 and pi/2. Show that

    d/dx arcsinx = 1/sqrt(1-x^2) with x between -1 and 1
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    Hello,

    Quote Originally Posted by SportfreundeKeaneKent View Post
    I don't even understand the wording of this. Also arcSINx?

    Denote by y=arcsinx, -1< or = x < or = 1, the inverse of y=sinx with x between -pi/2 and pi/2. Show that

    d/dx arcsinx = 1/sqrt(1-x^2) with x between -1 and 1
    arcsin x denotes the inverse function, as stated above.

    This means :

    y=\arcsin x \Longleftrightarrow x=\sin y.


    You want \frac{dy}{dx}.

    I'll do it the other way round :

    \frac{dx}{dy}=\frac{d}{dy}(\sin y)=\cos y=\pm \sqrt{1-\sin^2 y}

    Here, \cos y={\color{red}+}\sqrt{1-\sin^2 y}, because -pi/2<=y<=pi/2 --> cos y is positive.


    \frac{dx}{dy}=\sqrt{1-\sin^2y}=\sqrt{1-x^2}, because x=\sin y.


    Therefore \frac{d}{dx} \arcsin x=\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}


    Any question ?
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