How do you implicitly differentiate  y^2-xy-6=0?

Feb 16, 2016

You can do it like this:

Explanation:

${y}^{2} - x y - 6 = 0$

$\therefore D \left({y}^{2} - x y - 6\right) = D \left(0\right)$

$\therefore 2 y y ' - \left(x y ' + y\right) = 0$

$\therefore 2 y y ' - x y ' - y = 0$

$\therefore y ' \left(2 y - x\right) = y$

$y ' = \frac{y}{\left(2 y - x\right)}$