# How do you implicitly differentiate  y^2+(y-x)^2-y/x^2-3y?

Nov 26, 2017

First expand the brackets to give:
${y}^{2} + {\left(y - x\right)}^{2} - \frac{y}{x} ^ 2 - 3 y$
$= {y}^{2} + {y}^{2} - 2 x y + {x}^{2} - \frac{y}{x} ^ 2 - 3 y$
$= 2 {y}^{2} - 2 x y + {x}^{2} - y \times {x}^{-} 2 - 3 y$

By implicit differentiation (presumably with respect to x here):
$\frac{d}{\mathrm{dx}} {y}^{a} = a {y}^{a - 1} \times \frac{\mathrm{dy}}{\mathrm{dx}}$.
Also using the product rule:
If $f \left(x\right) = u v$, then $f ' \left(x\right) = u ' v + u v '$

So the original expression differentiates to:
$4 {y}^{1} \frac{\mathrm{dy}}{\mathrm{dx}} - 2 \left(x \frac{\mathrm{dy}}{\mathrm{dx}} + y \times 1\right) + 2 {x}^{1} - \left(- 2 {x}^{-} 3 y + {x}^{-} 2 \frac{\mathrm{dy}}{\mathrm{dx}}\right) - 3 {y}^{0} \frac{\mathrm{dy}}{\mathrm{dx}}$
$= 4 y \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x \frac{\mathrm{dy}}{\mathrm{dx}} - 2 y + 2 x + \frac{2 y}{x} ^ 3 - \frac{1}{x} ^ 2 \frac{\mathrm{dy}}{\mathrm{dx}}$