Question #3a275

Mar 12, 2017

See below

Explanation:

This is a classic example of implicit differentiation.
This is almost the same as taking the derivate with only one term $\left(\frac{d}{\mathrm{dx}}\right)$, but everytime there is a $y$, we write $\frac{\mathrm{dy}}{\mathrm{dx}}$, as we will be taking the derivative of "$y$ with respect to $x$." In other words, if there is a $y$ and an $x$ together, we must apply the product rule ($y$ will follow the same derivation as $x$).

The derivative of ${y}^{3} \cdot {e}^{x}$ is $\frac{d}{\mathrm{dx}} \left({y}^{3} \cdot {e}^{x}\right) = {y}^{3} \cdot {e}^{x} + 3 {y}^{2} {e}^{x} \frac{\mathrm{dy}}{\mathrm{dx}}$, by the product rule.

The derivative of $- {x}^{2}$ is $- 2 x$.

The derivative of $5 \cdot {e}^{y}$ is $5 \cdot {e}^{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$, as the $y$ is not with an $x$ term, so we cannot treat it as a constant.

And of course, the derivative of $1$ is $0$.

Now we can rearrange our equation so that the $\frac{\mathrm{dy}}{\mathrm{dx}}$ is isolated.
${y}^{3} \cdot {e}^{x} + 3 {y}^{2} {e}^{x} \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x + 5 \cdot {e}^{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$
$3 {y}^{2} {e}^{x} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + 5 \cdot {e}^{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 2 x - {y}^{3} \cdot {e}^{x}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x - {y}^{3} \cdot {e}^{x}}{3 {y}^{2} {e}^{x} + 5 \cdot {e}^{y}}$.

Hopefully I didn't do anything silly, and that this helps. Good luck :).