# How do you differentiate sqrt(e^(2x-2y))-xy^2=6?

Apr 26, 2016

Start by simplifying $\sqrt{{e}^{2 x - 2 y}}$

#### Explanation:

$\sqrt{{e}^{2 x - 2 y}} = {\left({e}^{2 x - 2 y}\right)}^{\frac{1}{2}} = {e}^{\left(\frac{1}{2}\right) \left(2 x - 2 y\right)} = {e}^{x - y}$

Rewrite:

$\sqrt{{e}^{2 x - 2 y}} - x {y}^{2} = 6$ as

${e}^{x - y} - x {y}^{2} = 6$.

Now differentiate term by term. Using the chain rule to differentiate ${e}^{x - y}$ and the product rule for $- x {y}^{2}$

$\frac{d}{\mathrm{dx}} \left({e}^{x - y}\right) - \frac{d}{\mathrm{dx}} \left(x {y}^{2}\right) = \frac{d}{\mathrm{dx}} \left(6\right)$.

${e}^{x - y} \frac{d}{\mathrm{dx}} \left(x - y\right) - \left[{y}^{2} + x 2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right] = 0$.

${e}^{x - y} \left(1 - \frac{\mathrm{dy}}{\mathrm{dx}}\right) - \left[{y}^{2} + 2 x y \frac{\mathrm{dy}}{\mathrm{dx}}\right] = 0$.

Now solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$. Distribute to remove brackets.

${e}^{x - y} - {e}^{x - y} \frac{\mathrm{dy}}{\mathrm{dx}} - {y}^{2} - 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$.

${e}^{x - y} - {y}^{2} = {e}^{x - y} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$.

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{x - y} - {y}^{2}}{{e}^{x - y} + 2 x y}$