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Math Help - Derivative Hellp meee!

  1. #1
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    Derivative Hellp meee!

    find f'(x) by applying this formula:
    f'(x_1) = \lim_{x\to x_1} \frac {f(x) - f(x_1)}{x - x_1}

    \csc x; x_1 = \frac {1}{2} \pi
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    find f'(x) by applying this formula:
    f'(x_1) = \lim_{x\to x_1} \frac {f(x) - f(x_1)}{x - x_1}

    \csc x; x_1 = \frac {1}{2} \pi
    I'm going to have to assume we know two limits:
    \lim_{h \to 0} \frac{1 - cos(h)}{h} = 0
    \lim_{h \to 0} \frac{sin(h)}{h} = 1

    (I'm going to use h instead of x_1 to save me a bit of coding.)

    f'(x) = \lim_{h \to 0} \frac{ csc(x+h) - csc(x) }{h}

    f'(x) = \lim_{h \to 0} \frac{ \frac{1}{sin(x+h)} - \frac{1}{sin(x)} }{h}

    f'(x) = \lim_{h \to 0} \frac{sin(x) - sin(x + h)}{hsin(x)sin(x+h)}

    Now expand the sin(x + h) = sin(x)cos(h) + sin(h)cos(x) in the numerator:
    f'(x) = \lim_{h \to 0} \frac{sin(x) - sin(x)cos(h) - sin(h)cos(x)}{hsin(x)sin(x+h)}

    f'(x) = \lim_{h \to 0} \frac{sin(x) - sin(x)cos(h)}{hsin(x)sin(x+h)} - \lim_{h \to 0} \frac{sin(h)cos(x)}{hsin(x)sin(x+h)}

    f'(x) = \lim_{h \to 0} \frac{sin(x)[1 - cos(h)]}{hsin(x)sin(x+h)} - \lim_{h \to 0} \frac{sin(h)}{h} \cdot \lim_{h \to 0}\frac{cos(x)}{sin(x)sin(x+h)}

    f'(x) = \lim_{h \to 0} \frac{1 - cos(h)}{h} \cdot \lim_{h \to 0} \frac{1}{sin(x+h)} - \lim_{h \to 0} \frac{sin(h)}{h} \cdot \lim_{h \to 0} \frac{cos(x)}{sin(x)sin(x+h)}

    You can now see the limits I mentioned above. So:
    f'(x) = 0 \cdot \lim_{h \to 0} \frac{1}{sin(x+h)} - 1 \cdot \lim_{h \to 0} \frac{cos(x)}{sin(x)sin(x+h)}

    f'(x) = -\lim_{h \to 0} \frac{cos(x)}{sin(x)sin(x+h)}

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

    -Dan
    Last edited by topsquark; December 3rd 2006 at 10:58 AM. Reason: Fixed the "-"
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  3. #3
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    There should be a minus after you expand the sine. That might lead to an error in the end.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    There should be a minus after you expand the sine. That might lead to an error in the end.
    I fixed it in the original post. Thanks for the catch!

    -Dan
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