1. ## Derivative

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

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

2. Originally Posted by deltaxray
Use the radius and tangent theorem to find the derivatives of:

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

$\displaystyle \frac{d}{dx} f(g(x))=g'(x)f'(g(x))$
3. 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 $\displaystyle f(x)=7-\sqrt{2x-x^2}$ is the lower half of the circle $\displaystyle (x-1)^2+(y-7)^2=1$, 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 $\displaystyle \frac{7-f(x)}{1-x}$. 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
$\displaystyle -\frac{1-x}{7-f(x)}=\frac{x-1}{7-f(x)}$. But the slope of the tangent line is f'(x), so one just plugs $\displaystyle f(x)=7-\sqrt{2x-x^2}$ into this $\displaystyle f'(x)=\frac{x-1}{7-f(x)}$ to get one's answer.