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Math Help - Differentiate questions

  1. #1
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    Differentiate questions

    Hi
    Can someone help me differentiate the following questions:
    1) \frac{sinx+cosx}{sinx-cosx}

    2)Find the local maximum and minimum values of f(x) for each of the following and state the corresponding x value of each. (Consider 0<= x <= 2\pi only):
    f(x)=2cosx+sin2x

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi
    Can someone help me differentiate the following questions:
    1) \frac{sinx+cosx}{sinx-cosx}

    2)Find the local maximum and minimum values of f(x) for each of the following and state the corresponding x value of each. (Consider 0<= x <= 2\pi only):
    f(x)=2cosx+sin2x

    P.S
    First, this should be in the Calculus sub-forum, not Pre-calculus.

    1) y = \frac{\sin{x} + \cos{x}}{\sin{x} - \cos{x}}.

    To differentiate this, use the quotient rule.

    \frac{dy}{dx} = \frac{(\sin{x} - \cos{x})\frac{d}{dx}(\sin{x} + \cos{x}) - (\sin{x} + \cos{x})\frac{d}{dx}(\sin{x} - \cos{x})}{(\sin{x} - \cos{x})^2}

     = \frac{(\sin{x} - \cos{x})(\cos{x} - \sin{x}) - (\sin{x} + \cos{x})(\cos{x} + \sin{x})}{(\sin{x} - \cos{x})^2}

     = \frac{-(\sin{x} - \cos{x})^2 - (\sin{x} + \cos{x})^2}{(\sin{x} - \cos{x})^2}

     = -1 - \left(\frac{\sin{x} + \cos{x}}{\sin{x} - \cos{x}}\right)^2.
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    Quote Originally Posted by Paymemoney View Post
    Hi
    Can someone help me differentiate the following questions:
    1) \frac{sinx+cosx}{sinx-cosx}

    2)Find the local maximum and minimum values of f(x) for each of the following and state the corresponding x value of each. (Consider 0<= x <= 2\pi only):
    f(x)=2cosx+sin2x

    P.S
    2) f(x) = 2\cos{x} + \sin{(2x)}

    f'(x) = -2\sin{x} + 2\cos{(2x)}.

    To find local maximums and minimums, let f'(x) = 0

    0 = -2\sin{x} + 2\cos{(2x)}

    0 = -2\sin{x} + 1 - 2\sin^2{x}

    2\sin^2{x} + 2\sin{x} - 1 = 0

    Let X = \sin{x}, so

    2X^2 + 2X - 1 = 0

    X = \frac{-2 \pm \sqrt{2^2 - 4(2)(-1)}}{2(2)}

     = \frac{-2 \pm \sqrt{12}}{4}

     = \frac{-2 \pm 2\sqrt{3}}{4}

     = \frac{-1 \pm \sqrt{3}}{2}.


    So \sin{x} = \frac{-1 + \sqrt{3}}{2} or \sin{x} = \frac{-1 - \sqrt{3}}{2}.

    Now get decimal approximations for x.
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    thanks for answering the questions.
    I thought this is pre-uni maths.
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  5. #5
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    It's clearly calculus. Therefore it goes in the calculus sub-forum.

    It's a bit misleading I know, but there you go. You should have also read the "Read this before posting" thread, where it is explained.
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