# How to find dy/dx from x^2y+3xy^2-x=5?

Mar 9, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 x y - 3 {y}^{2} + 1}{{x}^{2} + 6 x y}$

#### Explanation:

To solve this, we need to use implicit differentiation.

Since we are finding the derivative in respect to $x$, anytime we find the derivative of $y$, we need to tag on a $\frac{\mathrm{dy}}{\mathrm{dx}}$ after.

To start off, we need to use the Product Rule for the first two terms, since there are two variables being multiplied together.

The Product Rule states:

$f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

So we can now set it up like this:

${x}^{2} y + 3 x {y}^{2} - x = 5$

$2 x y + {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 3 {y}^{2} + 6 x y \frac{\mathrm{dy}}{\mathrm{dx}} - 1 = 0$

Now that we have the derivative, we need to clean up the answer a bit. We need to move each term without the $\frac{\mathrm{dy}}{\mathrm{dx}}$ to the other side. So we get:

${x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 6 x y \frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x y - 3 {y}^{2} + 1$

Now we factor out a $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} \left({x}^{2} + 6 x y\right) = - 2 x y - 3 {y}^{2} + 1$

Divide:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 x y - 3 {y}^{2} + 1}{{x}^{2} + 6 x y}$

Mar 9, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - 2 x y - 3 {y}^{2}}{{x}^{2} + 6 x y}$

#### Explanation:

We have:

${x}^{2} y + 3 x {y}^{2} - x = 5$

Differentiating wrt $x$ by applying the product rule we get:

${x}^{2} \left(\frac{d}{\mathrm{dx}} y\right) + \left(\frac{d}{\mathrm{dx}} {x}^{2}\right) y + \left(3 x\right) \left(\frac{d}{\mathrm{dx}} {y}^{2}\right) + \left(\frac{d}{\mathrm{dx}} 3 x\right) {y}^{2} - \frac{d}{\mathrm{dx}} x = \frac{d}{\mathrm{dx}} 5$

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

Then we apply the chain rule to perform the implicit differentiation:

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

Now, we collect terms and solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\left({x}^{2} + 6 x y\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 1 - 2 x y - 3 {y}^{2}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - 2 x y - 3 {y}^{2}}{{x}^{2} + 6 x y}$