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Thread: 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/$\displaystyle sqrt(1-x^2)$
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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/$\displaystyle sqrt(1-x^2)$
    the domain of the function $\displaystyle \sin x$ is the range of the inverse function $\displaystyle \sin ^{-1} x$ and the range of the function $\displaystyle \sin x $ is the domain of the function $\displaystyle \sin ^{-1}x$ so

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


    next question

    show that

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

    let

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

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

    now derive

    $\displaystyle dy (\cos y ) = dx$

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

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


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

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

    but sin y =x so

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

    $\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$
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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/$\displaystyle sqrt(1-x^2)$
    $\displaystyle y = \arcsin x \Rightarrow x = \sin y$.

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

    Your job is to get $\displaystyle \cos y$ in terms of $\displaystyle x$ 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 $\displaystyle \sin x$ is the range of the inverse function $\displaystyle \sin ^{-1} x$ and the range of the function $\displaystyle \sin x $ is the domain of the function $\displaystyle \sin ^{-1}x$ so

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


    next question

    show that

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

    let

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

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

    now derive

    $\displaystyle dy (\cos y ) = dx$

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

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


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

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

    but sin y =x so

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

    $\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$
    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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