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Inverse Cos Derivative

  1. Jun 27, 2006 #1
    Yeah, I tried doing this question and just wanted to check if I was correct.

    Write down the derivative of [tex]\arccos (4x^2)[/tex] and state the domain for which the derivative applies.

    [tex]\mbox{let}\ y = \arccos (4x^2), x\ \epsilon\ [-1, 1], y\ \epsilon\ [0, \pi][/tex]
    [tex]\Leftrightarrow x = \frac{\sqrt{\cos y}}{2}[/tex]

    [tex]\begin{align*}
    \frac{d}{dx} x = \frac{d}{dx} \frac{\sqrt{\cos y}}{2} \\
    1 = \frac{-\sin y}{4 \sqrt{\cos y}} \cdot \frac{dy}{dx} \\
    \frac{dy}{dx} = \frac{4 \sqrt{\cos y}}{-\sin y}
    \end{align*}[/tex]

    [tex]\cos^2 y + \sin^2 y = 1[/tex]
    [tex]\sin y = \sqrt{1 - \cos^2 y}, \mbox{because}\ y\ \epsilon\ [0, \pi]\ \mbox{so}\ \sin y \geq 0[/tex]

    [tex]\therefore \frac{dy}{dx} = \frac{4\sqrt{\cos y}}{-\sqrt{1 - \cos^2 y}}[/tex]
     
    Last edited: Jun 27, 2006
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  3. Jun 27, 2006 #2

    HallsofIvy

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    Don't forget that cos y= 4x2 so
    [tex]\frac{dy}{dx}= -\frac{8x}{\sqrt{1- 16x^2}}[/tex]
    I suspect your teacher will prefer that form.
     
  4. Jun 27, 2006 #3
    Thanks, and I think it should be
    [tex]\frac{dy}{dx} = -\frac{8x}{\sqrt{1 - 16x^4}}[/tex]
     
  5. Jun 27, 2006 #4

    George Jones

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    What about the domain of validity. Note that with with respect to this, there is a mistake in line 3 (first line of mathematics) in post #1.
     
  6. Jun 27, 2006 #5
    The domain would be [tex]x\ \epsilon\ (-\frac{1}{4},\ \frac{1}{4})[/tex] simply because when x = 0.25, the denom is undefined.

    And what is the mistake in line 3?
     
  7. Jun 27, 2006 #6

    George Jones

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    The argument of arccos is an element of the closed interval [itex]\left[ -1 , 1 \right][/itex], i.e., [itex]4x^{2} \epsilon \left[ -1 , 1 \right][/itex], not [itex]x[/itex]. This immediately gives "almost" the domain of validity.
     
    Last edited: Jun 27, 2006
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