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

Nov 29, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 x {e}^{{x}^{2} - y}}{3 {y}^{2} + 4 x y + {y}^{- 2} - {e}^{{x}^{2} - y}}$

#### Explanation:

When given a function in terms of $x$ and $y$, and we want to find $\frac{\mathrm{dy}}{\mathrm{dx}}$ in terms of $x$ and $y$, we implicity differentiate.

Using our equation, we can differentiate in terms of $x$:
$\frac{d}{\mathrm{dx}} \left[{y}^{3}\right] + \frac{d}{\mathrm{dx}} \left[{x}^{2}\right] \frac{d}{\mathrm{dx}} \left[{y}^{2}\right] = \frac{d}{\mathrm{dx}} \left[{y}^{- 1}\right] - \frac{d}{\mathrm{dx}} \left[{e}^{{x}^{2} - y}\right]$

$= \frac{d}{\mathrm{dx}} \left[{y}^{3}\right] + 2 x \frac{d}{\mathrm{dx}} \left[{y}^{2}\right] = \frac{d}{\mathrm{dx}} \left[{y}^{- 1}\right] - \left(\frac{d}{\mathrm{dx}} \left[{x}^{2}\right] - \frac{d}{\mathrm{dx}} \left[y\right]\right) {e}^{{x}^{2} - y}$

$= \frac{d}{\mathrm{dx}} \left[{y}^{3}\right] + 2 x \frac{d}{\mathrm{dx}} \left[{y}^{2}\right] = \frac{d}{\mathrm{dx}} \left[{y}^{- 1}\right] - \left(2 x - \frac{d}{\mathrm{dx}} \left[y\right]\right) {e}^{{x}^{2} - y}$

The chain rule gives us: $\frac{d}{\mathrm{dx}} = \frac{d}{\mathrm{dy}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{d}{\mathrm{dy}} \left[{y}^{3}\right] \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x \frac{d}{\mathrm{dy}} \left[{y}^{2}\right] \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dy}} \left[{y}^{- 1}\right] \frac{\mathrm{dy}}{\mathrm{dx}} - \left(2 x - \frac{d}{\mathrm{dy}} \left[y\right] \frac{\mathrm{dy}}{\mathrm{dx}}\right) {e}^{{x}^{2} - y}$

By a process of rearranging and simplifying:
$3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = - {y}^{- 2} \frac{\mathrm{dy}}{\mathrm{dx}} - \left(2 x - 1 \frac{\mathrm{dy}}{\mathrm{dx}}\right) {e}^{{x}^{2} - y}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} \left[3 {y}^{2} + 4 x y\right] = - {y}^{- 2} \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x {e}^{{x}^{2} - y} + \frac{\mathrm{dy}}{\mathrm{dx}} \left[{e}^{{x}^{2} - y}\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left[3 {y}^{2} + 4 x y\right] = \frac{\mathrm{dy}}{\mathrm{dx}} \left[- {y}^{- 2} + {e}^{{x}^{2} - y}\right] - 2 x {e}^{{x}^{2} - y}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left[3 {y}^{2} + 4 x y + {y}^{- 2} - {e}^{{x}^{2} - y}\right] = - 2 x {e}^{{x}^{2} - y}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 x {e}^{{x}^{2} - y}}{3 {y}^{2} + 4 x y + {y}^{- 2} - {e}^{{x}^{2} - y}}$