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Math Help - calculus Identities

  1. #1
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    calculus Identities

    define the function arcsin(x), stating for which values of x your function is defined. Explained why the derivative of arcsin(x) is 1/ sqrt(1-x^2)
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    Quote Originally Posted by manalive04 View Post
    define the function arcsin(x), stating for which values of x your function is defined. Explained why the derivative of arcsin(x) is 1/ sqrt(1-x^2)
    the domain of the function \sin x is the range of the inverse function \sin ^{-1} x and the range of the function \sin x is the domain of the function \sin ^{-1}x so

    the values of x which \sin ^{-1}x is the defined is the domain of the function \sin ^{-1}x and the domain of this function is the range of the function \sin x so the values of x which \sin ^{-1}x is defined is [-1,1]


    next question

    show that

    \frac{d}{dx}\left(\sin  ^{-1}x\right)=\frac{1}{\sqrt{1-x^2}}

    let

    y=\sin ^{-1}x want to find \frac{dy}{dx}

    \sin y = x right ?? take sin for the both sides

    now derive

    dy (\cos y ) = dx

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

    \left(\frac{dy}{dx}\right)^2 = \left(\frac{1}{\cos y}\right)^2


    \left(\frac{dy}{dx}\right)^2 = \left(\frac{1}{\cos ^2 y}\right)

    \left(\frac{dy}{dx}\right)^2 = \left(\frac{1}{1-\sin ^2 y}\right)

    but sin y =x so

    \left(\frac{dy}{dx}\right)^2 = \left(\frac{1}{1-x^2}\right)

    \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}
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  3. #3
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    Quote Originally Posted by manalive04 View Post
    define the function arcsin(x), stating for which values of x your function is defined. Explained why the derivative of arcsin(x) is 1/ sqrt(1-x^2)
    y = \arcsin x \Rightarrow x = \sin y.

    Therefore \frac{dx}{dy} = \cos y \Rightarrow \frac{dy}{dx} = \frac{1}{\cos y} = ....

    Your job is to get \cos y in terms of x and hence fill in the ....
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  4. #4
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    Quote Originally Posted by Amer View Post
    the domain of the function \sin x is the range of the inverse function \sin ^{-1} x and the range of the function \sin x is the domain of the function \sin ^{-1}x so

    the values of x which \sin ^{-1}x is the defined is the domain of the function \sin ^{-1}x and the domain of this function is the range of the function \sin x so the values of x which \sin ^{-1}x is defined is [-1,1]


    next question

    show that

    \frac{d}{dx}\left(\sin ^{-1}x\right)=\frac{1}{\sqrt{1-x^2}}

    let

    y=\sin ^{-1}x want to find \frac{dy}{dx}

    \sin y = x right ?? take sin for the both sides

    now derive

    dy (\cos y ) = dx

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

    \left(\frac{dy}{dx}\right)^2 = \left(\frac{1}{\cos y}\right)^2


    \left(\frac{dy}{dx}\right)^2 = \left(\frac{1}{\cos ^2 y}\right)

    \left(\frac{dy}{dx}\right)^2 = \left(\frac{1}{1-\sin ^2 y}\right)

    but sin y =x so

    \left(\frac{dy}{dx}\right)^2 = \left(\frac{1}{1-x^2}\right)

    \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}
    I'm glad you left the rejection of the negative root as something for the OP to think about and justify.
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