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Thread: Double Check

  1. #1
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    Exclamation Double Check

    Hi

    Could someone check to c if these are the right answers:


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

    $\displaystyle \frac{d}{dx}*\frac{1}{arcsinx} = \frac{-1}{(arcsin)^2*\sqrt{1-x^2}}$

    If they're wrong, could someone please show me the right way to do them??
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  2. #2
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    Quote Originally Posted by xwrathbringerx View Post
    Hi

    Could someone check to c if these are the right answers:


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

    should be $\displaystyle \textcolor{red}{\frac{1}{\sqrt{1-x^2}}}$

    $\displaystyle \frac{d}{dx}*\frac{1}{\arcsin{x}} = \frac{-1}{(\arcsin{\textcolor{red}{x}})^2*\sqrt{1-x^2}}$

    If they're wrong, could someone please show me the right way to do them??
    ...
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  3. #3
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    Ummm what's wrong with the second answer?
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  4. #4
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    Quote Originally Posted by xwrathbringerx View Post
    Hi

    $\displaystyle \frac{d}{dx}*\frac{1}{arcsinx} = \frac{-1}{(arcsin)^2*\sqrt{1-x^2}}$
    arcsin(of what?) in the denominator
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  5. #5
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    it's arcsin of x.

    Sori
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  6. #6
    Senior Member I-Think's Avatar
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    $\displaystyle y=tan^{'}f(x)$

    $\displaystyle tan y=f(x)$
    $\displaystyle \frac{d}{dx} tan y= \frac{d}{dx} f(x)$

    $\displaystyle sec^2y\frac{dy}{dx}=f^{'}(x)$
    $\displaystyle (tan^2y+1)\frac{dy}{dx}=f^{'}(x)$
    $\displaystyle \frac{dy}{dx}=\frac{f^{'}(x)}{[f(x)]^2+1}$

    Find $\displaystyle \frac{d}{dx} arctan(\frac{x}{\sqrt{1-x^2}})$

    If $\displaystyle g(x)=\frac{x}{\sqrt{1-x^2}}$
    Then $\displaystyle g^{'}x=\frac{\sqrt{1-x^2}+x^2(1-x^2)^{-0.5}}{1-x^2}$
    And $\displaystyle [g(x)]^2=\frac{x^2}{1-x^2}$

    So
    $\displaystyle \frac{d}{dx} arctan(\frac{x}{\sqrt{1-x^2}})=\frac{\frac{\sqrt{1-x^2}+x^2(1-x^2)^{-0.5}}{1-x^2}}{\frac{x^2}{1-x^2}+1}$
    Which becomes
    $\displaystyle \sqrt{1-x^2}+\frac{x^2}{\sqrt{1-x^2}}$
    Becoming
    $\displaystyle \frac{1}{\sqrt{1-x^2}}$

    If you're wondering how to do it.
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