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Math Help - Derivative

  1. #1
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    Derivative

    Use the radius and tangent theorem to find the derivatives of:

    f(x)=7-\sqrt{2x-x^2}

    Thanks in advance
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by deltaxray View Post
    Use the radius and tangent theorem to find the derivatives of:

    f(x)=7-\sqrt{2x-x^2}

    Thanks in advance
    I'm sorry but Google returns no hits for "radius and tangent theorem". This problem needs only the chain rule:

    \frac{d}{dx} f(g(x))=g'(x)f'(g(x))

    CB
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  3. #3
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    I believe that the "radius and tangent theorem" is the theorem from geometry that a line tangent to a circle is perpendicular to the radius of the circle to the point of tangency. Since the function f(x)=7-\sqrt{2x-x^2} is the lower half of the circle (x-1)^2+(y-7)^2=1, the circle of radius 1 and center (1,7), a point (x,f(x)) is the endpoint of a radius whose other endpoint is the center (1,7), and thus with slope \frac{7-f(x)}{1-x}. Therefore the line tangent to our circle at that point is perpendicular to that radius, and so has opposite reciprocal slope; the tangent line at (x,f(x)) has slope
    -\frac{1-x}{7-f(x)}=\frac{x-1}{7-f(x)}. But the slope of the tangent line is f'(x), so one just plugs f(x)=7-\sqrt{2x-x^2} into this f'(x)=\frac{x-1}{7-f(x)} to get one's answer.

    --Kevin C.
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