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Math Help - differential

  1. #1
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    differential

    y=sinx+cosx is a function defined in the interval [0,pi/6], find derivative of f^-1(x)?
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  2. #2
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    Re: differential

    The derivative of f^{-1}(x) is \frac{1}{f'(x)}.

    - Hollywood
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  3. #3
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    Re: differential

    Yes it is but I couldn't find the answer given by this method.
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  4. #4
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    Re: differential

    I couldn't find the answer. Answer is 1/sqrt(2-x^2)

    y^2=1+sin2x y=sqrt(1+sin2x) and if we differentiate y, and write it

    1/dy/dx, we find 1/cosx/sqrt(1+sin2x).
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  5. #5
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    Re: differential

    You have y=\sin{x}+\cos{x} so \frac{dy}{dx} should be \cos{x}-\sin{x}. I don't see how you get \frac{1}{\sqrt{2-x^2}}. Are you sure the problem and answer are both copied correctly?

    I follow you on the second line of your post. On the third line you should have \cos{2x} instead of \cos{x} giving \frac{\sqrt{1+\sin{2x}}}{\cos{2x}}.

    - Hollywood
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  6. #6
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    Re: differential

    [QUOTE=hollywood;787266]You have y=\sin{x}+\cos{x} so \frac{dy}{dx} should be \cos{x}-\sin{x}. I don't see how you get \frac{1}{\sqrt{2-x^2}}. Are you sure the problem and answer are both copied correctly?

    I follow you on the second line of your post. On the third line you should have \cos{2x} instead of \cos{x} giving \frac{\sqrt{1+\sin{2x}}}{\cos{2x}}.
    y^2=1+sin2x, f^-1(x)=1/2.arcsin(x^2-1),x/(2x^2-x^4)
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  7. #7
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    Re: differential

    Oh, sure enough. It should be \frac{x}{\sqrt{2x^2-x^4}}, and then you can cancel x giving \frac{1}{\sqrt{2-x^2}}.

    In my solution I forgot to swap y and x, so I should have had \cos{y}-\sin{y}, which gives the same thing if you substitute y=\frac{1}{2}\sin^{-1}(x^2-1) (after a lot of algebra). Your solution is more direct. Well done.

    - Hollywood
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