# How do you implicitly differentiate -1=xy^2+2x^2y-e^ycsc(3x/y) ?

Jun 19, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\frac{3 {e}^{y}}{y} \csc \left(\frac{3 x}{y}\right) \cot \left(\frac{3 x}{y}\right) + {y}^{2} + 4 x y}{2 x y + 2 {x}^{2} - {e}^{y} \csc \left(\frac{3 x}{y}\right) + \frac{3 x}{y} ^ 2 \csc \left(\frac{3 x}{y}\right) \cot \left(\frac{3 x}{y}\right)}$

#### Explanation:

As $- 1 = x {y}^{2} + 2 {x}^{2} y - {e}^{y} \csc \left(3 \frac{x}{y}\right)$, taking differential of each side

${y}^{2} + 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} + 4 x y + 2 {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - \left({e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} \csc \left(\frac{3 x}{y}\right) - {e}^{y} \csc \left(\frac{3 x}{y}\right) \cot \left(\frac{3 x}{y}\right) \left(\frac{3}{y} - \frac{3 x}{y} ^ 2 \frac{\mathrm{dy}}{\mathrm{dx}}\right)\right) = 0$

or $\left(2 x y + 2 {x}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{\mathrm{dy}}{\mathrm{dx}} \left({e}^{y} \csc \left(\frac{3 x}{y}\right) - \frac{3 x}{y} ^ 2 \csc \left(\frac{3 x}{y}\right) \cot \left(\frac{3 x}{y}\right)\right) + \frac{3 {e}^{y}}{y} \csc \left(\frac{3 x}{y}\right) \cot \left(\frac{3 x}{y}\right) = - {y}^{2} - 4 x y$

or $\left(2 x y + 2 {x}^{2} - {e}^{y} \csc \left(\frac{3 x}{y}\right) + \frac{3 x}{y} ^ 2 \csc \left(\frac{3 x}{y}\right) \cot \left(\frac{3 x}{y}\right)\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - \left[\frac{3 {e}^{y}}{y} \csc \left(\frac{3 x}{y}\right) \cot \left(\frac{3 x}{y}\right) + {y}^{2} + 4 x y\right\}$

or $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\frac{3 {e}^{y}}{y} \csc \left(\frac{3 x}{y}\right) \cot \left(\frac{3 x}{y}\right) + {y}^{2} + 4 x y}{2 x y + 2 {x}^{2} - {e}^{y} \csc \left(\frac{3 x}{y}\right) + \frac{3 x}{y} ^ 2 \csc \left(\frac{3 x}{y}\right) \cot \left(\frac{3 x}{y}\right)}$