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Math Help - differentiation of trig functions in mod

  1. #1
    Junior Member cupid's Avatar
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    Question differentiation of trig functions in mod

    The Question and My Attempt in the pic

    i got correct answer but i think my method has some flaw ...

    Please someone verify.
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  2. #2
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    It's fine.
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  3. #3
    Junior Member cupid's Avatar
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    Can you help me to do this one ...
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  4. #4
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    You will need to consider how you can rewrite \displaystyle |x| and \displaystyle |\sin{x}| in the neighbourhood of \displaystyle x = -\frac{\pi}{4}.
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  5. #5
    Junior Member cupid's Avatar
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    I gues both |x| and |sin x| will be negative so |x| = -x and |sin x| = -sin x

    the i can take log on f(x)

    that way i'm getting something weird .... like (pi/4)^(1/√2) * [4/√2 pi]
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  6. #6
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    I get \left(\dfrac{\pi}{4}\right)^{\dfrac{1}{\sqrt{2}}}\  left(-\dfrac{\ln{\dfrac{\pi}{4}}}{\sqrt{2}} - \dfrac{2\sqrt{2}}{\pi}\right).

    f(x) = \left(-x\right)^{-\sin{x}} = e^{\ln{\left(-x\right)^{-\sin{x}}}} = e^{-\sin{x}\ln{\left(-x\right)}}

    Let u = -\sin{x}\ln{\left(-x\right)}. Then, du = -\cos{x}\ln{\left(-x\right)} - \dfrac{\sin{x}}{x}.

    Substitute u = -\sin{x}\ln{\left(-x\right)} in f(x).

    f(x) = e^u.

    Then, f'(x) is easy to compute.

    f'(x) = e^u du = e^{-\sin{x}\ln{\left(-x\right)}} \left(-\cos{x}\ln{\left(-x\right)} - \dfrac{\sin{x}}{x}\right) = \left(-x\right)^{-\sin{x}} \left(-\cos{x}\ln{\left(-x\right)} - \dfrac{\sin{x}}{x}\right)

    Substitute x = -\dfrac{\pi}{4} in f'(x).

    f'(-\dfrac{\pi}{4}) = \left(--\dfrac{\pi}{4}\right)^{-\sin{\left(-\dfrac{\pi}{4}\right)}} \left(-\cos{\left(-\dfrac{\pi}{4}\right)}\ln{\left(--\dfrac{\pi}{4}\right)} - \dfrac{\sin{\left(-\dfrac{\pi}{4}\right)}}{-\dfrac{\pi}{4}}\right)

    Get rid of the double negatives.

    f'(-\dfrac{\pi}{4}) = \left(\dfrac{\pi}{4}\right)^{-\sin{\left(-\dfrac{\pi}{4}\right)}} \left(-\cos{\left(-\dfrac{\pi}{4}\right)}\ln{\left(\dfrac{\pi}{4}\rig  ht)} + \dfrac{\sin{\left(-\dfrac{\pi}{4}\right)}}{\dfrac{\pi}{4}}\right)

    Next, get rid of the sines and cosines: sin{\left(-\dfrac{\pi}{4}\right)} = -\dfrac{1}{\sqrt{2}} and cos{\left(-\dfrac{\pi}{4}\right)} = \dfrac{1}{\sqrt{2}}.

    f'(-\dfrac{\pi}{4}) = \left(\dfrac{\pi}{4}\right)^{--\dfrac{1}{\sqrt{2}}} \left(-\dfrac{1}{\sqrt{2}}\ln{\left(\dfrac{\pi}{4}\right)  } + \dfrac{-\dfrac{1}{\sqrt{2}}}{\dfrac{\pi}{4}}\right)

    Get rid of double negatives.

    f'(-\dfrac{\pi}{4}) = \left(\dfrac{\pi}{4}\right)^{\dfrac{1}{\sqrt{2}}} \left(-\dfrac{1}{\sqrt{2}}\ln{\left(\dfrac{\pi}{4}\right)  } + \dfrac{-\dfrac{1}{\sqrt{2}}}{\dfrac{\pi}{4}}\right)

    Get rid of the fraction within a fraction within a fraction within a fraction.

    f'(-\dfrac{\pi}{4}) = \left(\dfrac{\pi}{4}\right)^{\dfrac{1}{\sqrt{2}}} \left(-\dfrac{1}{\sqrt{2}}\ln{\left(\dfrac{\pi}{4}\right)  } - \dfrac{4}{\sqrt{2}\pi}\right) = \left(\dfrac{\pi}{4}\right)^{\dfrac{1}{\sqrt{2}}} \left(-\dfrac{1}{\sqrt{2}}\ln{\left(\dfrac{\pi}{4}\right)  } - \dfrac{4\sqrt{2}}{2\pi}\right) = \left(\dfrac{\pi}{4}\right)^{\dfrac{1}{\sqrt{2}}} \left(-\dfrac{1}{\sqrt{2}}\ln{\left(\dfrac{\pi}{4}\right)  } - \dfrac{2\sqrt{2}}{\pi}\right)

    Edit: Prove It's method seems less convoluted. :P
    Last edited by NOX Andrew; February 27th 2011 at 07:00 AM.
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  7. #7
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    Well let's see...

    I agree that you would rewrite this as \displaystyle y = (-x)^{-\sin{x}}.

    So \displaystyle \ln{y} = \ln{\left[(-x)^{-\sin{x}}\right]}

    \displaystyle \ln{y} = -\sin{x}\ln{(-x)}

    \displaystyle \frac{d}{dx}(\ln{y}) = \frac{d}{dx}\left[-\sin{x}\ln{(-x)}\right]

    \displaystyle \frac{1}{y}\,\frac{dy}{dx} = -\frac{\sin{x}}{x} - \cos{x}\ln{(-x)}

    \displaystyle \frac{1}{(-x)^{-\sin{x}}}\,\frac{dy}{dx} = -\frac{\sin{x}}{x} - \cos{x}\ln{(-x)}

    \displaystyle \frac{dy}{dx} = (-x)^{-\sin{x}}\left[-\frac{\sin{x}}{x} - \cos{x}\ln{(-x)}\right].


    When this is evaluated at \displaystyle x = -\frac{\pi}{4}

    \displaystyle \frac{dy}{dx} = \left(\frac{\pi}{4}\right)^{\frac{\sqrt{2}}{2}}\le  ft[-\frac{2\sqrt{2}}{\pi} - \frac{\sqrt{2}}{2}\ln{\left(\frac{\pi}{4}\right)}\  right].
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  8. #8
    Junior Member cupid's Avatar
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    Hi guys ...

    really thanks for the replies ... i really appreciate it ......

    but the answer is (a)

    i am not getting how it will be 4/pi and how will the log term be positive ???

    really thanks for help
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  9. #9
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    \displaystyle -\frac{\sqrt{2}}{2}\ln{\left(\frac{\pi}{4}\right)} = \frac{\sqrt{2}}{2}\ln{\left[\left(\frac{\pi}{4}\right)^{-1}\right]} = \frac{\sqrt{2}}{2}\ln{\left(\frac{4}{\pi}\right)}
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  10. #10
    Junior Member cupid's Avatar
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    OH YES!!!

    That was dumb of me

    sorry for trouble

    Thanks "prove it" for help
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