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

Aug 23, 2016

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

#### Explanation:

Performing implicit differentiation involves differentiating each term on its own. We assume that $y$ is a function of $x$ so that any time we differentiate a function of $y$, we will still have our unknown $\frac{\mathrm{dy}}{\mathrm{dx}}$ term left over.

$\frac{d}{\mathrm{dx}} 2 = \textcolor{red}{\frac{d}{\mathrm{dx}} \left({e}^{x {y}^{3} - x}\right)} - \textcolor{g r e e n}{\frac{d}{\mathrm{dx}} \left({y}^{2} {x}^{3}\right)} + \frac{d}{\mathrm{dx}} \left(y\right)$

The first term is a constant, therefore the derivative is $0$. The last term is simply $\frac{\mathrm{dy}}{\mathrm{dx}}$. For clarity, lets go over the other two terms individually.

For the $\textcolor{red}{\text{red}}$ term, we can use the fact that $\frac{d}{\mathrm{dx}} \left({e}^{f} \left(x\right)\right) = {e}^{f} \left(x\right) f ' \left(x\right)$.

$\textcolor{red}{\frac{d}{\mathrm{dx}} \left({e}^{x {y}^{3} - x}\right) = \left({e}^{x {y}^{3} - x}\right) \frac{d}{\mathrm{dx}} \left(x {y}^{3} - x\right)}$

$\textcolor{red}{= \left({e}^{x {y}^{3} - x}\right) \left(\frac{d}{\mathrm{dx}} \left(x {y}^{3}\right) - \frac{d}{\mathrm{dx}} \left(x\right)\right)}$

Apply the product rule, $\frac{d}{\mathrm{dx}} f \left(x\right) g \left(x\right) = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$.

$\textcolor{red}{= \left({e}^{x {y}^{3} - x}\right) \left(\frac{d}{\mathrm{dx}} \left(x\right) \cdot {y}^{3} + x \cdot \frac{d}{\mathrm{dx}} \left({y}^{3}\right) - 1\right)}$

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

The green term is easier, we just need to apply the product rule again.

$\textcolor{g r e e n}{\frac{d}{\mathrm{dx}} \left({y}^{2} {x}^{3}\right) = \frac{d}{\mathrm{dx}} \left({y}^{2}\right) \cdot {x}^{3} + {y}^{2} \frac{d}{\mathrm{dx}} \left({x}^{3}\right)}$

$\textcolor{g r e e n}{= 2 y \frac{\mathrm{dy}}{\mathrm{dx}} \cdot {x}^{3} + 3 {y}^{2} {x}^{2}}$

Now we can put everything back together. Don't forget that the middle term was negative.

$0 = {e}^{\left(x {y}^{3} - x\right)} \left({y}^{3} + 3 x {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - 1\right) - 2 y \frac{\mathrm{dy}}{\mathrm{dx}} \cdot {x}^{3} - 3 {y}^{2} {x}^{2} + \frac{\mathrm{dy}}{\mathrm{dx}}$

Separate the terms that have $\frac{\mathrm{dy}}{\mathrm{dx}}$ and move them to the left hand side.

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

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

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$.

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