Originally Posted by
MatteNoob Mr Fantastic
Oh, sorry about the typo.
So if I want to write the equation with respect to y, I could do this:
$\displaystyle (x-3)^2 +(y+1)^2 = 25$
$\displaystyle (y+1)^2 = 5^2 - (x-3)^2 $
$\displaystyle y = \pm \sqrt{5^2 - (x-3)^2}-1$
Then, if I want the derivative, I could;
$\displaystyle u = -x^2 + 6x {\color{red}+} 16 \;\;\;\; u\prime = -2x +6$
$\displaystyle y\prime = \pm \left(\sqrt{u}\right)\prime \cdot u\prime - (1)\prime$
So that
$\displaystyle y\prime = \pm \frac{-2x+6}{2\sqrt{-x^2+6x {\color{red}+} 16}} = \underline{\underline{ \pm \frac{3-x}{\sqrt{6x - x^2 {\color{red}+} 16}}}}$
Edit: Corrected a typo.