How do you implicitly differentiate -y^2=e^(2x-4y)-2yx ?

May 1, 2018

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

Explanation:

We can write this as:
$2 y x - {y}^{2} = {\left({e}^{x - 2 y}\right)}^{2}$

Now we take $\frac{d}{\mathrm{dx}}$ of each term:
$\frac{d}{\mathrm{dx}} \left[2 y x\right] - \frac{d}{\mathrm{dx}} \left[{y}^{2}\right] = \frac{d}{\mathrm{dx}} \left[{\left({e}^{x - 2 y}\right)}^{2}\right]$

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

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

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

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

Using the chain rule we get:
$\frac{d}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{d}{\mathrm{dy}}$

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

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

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

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

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

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