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

Nov 7, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x {e}^{x + y}}{1 + 2 {e}^{x + y} + {y}^{2} {e}^{x + y}}$

Explanation:

This explanation is going to be long by the nature of implicit differentiation so please bear with me on this one.

The first thing to do is take the derivative with respect to $x$ of both sides: $\frac{d}{\mathrm{dx}} \left[y\right] = \frac{d}{\mathrm{dx}} \left[\left(x - {y}^{2}\right) {e}^{x + y}\right]$. For this, we must use the product rule, which states: $\frac{d}{\mathrm{dx}} \left[f \left(x\right) g \left(x\right)\right] = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$.

Since this problem has many pieces to it, I am going to take this part of the problem in pieces. Let's assign $f \left(x\right)$ and $g \left(x\right)$. $f \left(x\right) = \left(x - {y}^{2}\right) \mathmr{and} g \left(x\right) = {e}^{x + y}$. Now to find their derivatives: $f ' \left(x\right) = 1 - 2 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \mathmr{and} g ' \left(x\right) = \left(1 + \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\right) {e}^{x + y}$. I used the chain rule for $g ' \left(x\right)$. $g ' \left(x\right)$ can be rewritten as $\left({e}^{x + y} + {e}^{x + y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\right)$.

Now to plug in: $\frac{\mathrm{dy}}{\mathrm{dx}} = \left(1 - 2 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\right) \left({e}^{x + y}\right) + \left(x - {y}^{2}\right) \left({e}^{x + y} + {e}^{x + y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\right)$. Next, we distribute the two products to get: $\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x + y} - 2 y {e}^{x + y} \frac{\mathrm{dy}}{\mathrm{dx}} + x {e}^{x + y} - {y}^{2} {e}^{x + y} \frac{\mathrm{dy}}{\mathrm{dx}}$. The next thing to do is gather all the terms containing $\frac{\mathrm{dy}}{\mathrm{dx}}$onto the same side.
$\frac{\mathrm{dy}}{\mathrm{dx}} + 2 y {e}^{x + y} \frac{\mathrm{dy}}{\mathrm{dx}} + {y}^{2} {e}^{x + y} \frac{\mathrm{dy}}{\mathrm{dx}} = x {e}^{x + y}$.

Now, on the left side we factor out $\frac{\mathrm{dy}}{\mathrm{dx}}$ from all terms to get $\frac{\mathrm{dy}}{\mathrm{dx}} \left(1 + 2 y {e}^{x + y} + {y}^{2} {e}^{x + y}\right) = x {e}^{x + y}$. Finally, to get $\frac{\mathrm{dy}}{\mathrm{dx}}$ by itself, we divide both sides by$\left(1 + 2 y {e}^{x + y} + {y}^{2} {e}^{x + y}\right)$ to get our final answer of $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x {e}^{x + y}}{1 + 2 y {e}^{x + y} + {y}^{2} {e}^{x + y}}$.