# What is the value of the implicit derivative of y=y^2x^2 at x=3?

Dec 13, 2015

$0 , - \frac{2}{27}$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left[y = {y}^{2} {x}^{2}\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2} \frac{d}{\mathrm{dx}} \left[{y}^{2}\right] + {y}^{2} \frac{d}{\mathrm{dx}} \left[{x}^{2}\right]$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} - 2 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x {y}^{2}$

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

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

Now that we have the implicit derivative, we must find the point or points when $x = 3$, so we can plug them in to find the value(s) of the implicit derivative.

We know that $x = 3$, so:

$y = {y}^{2} {x}^{2}$
$y = 9 {y}^{2}$

With this:

$0 = 9 {y}^{2} - y$
$0 = y \left(9 y - 1\right)$

$y = 0 , \frac{1}{9}$

We now have to two points $\left(3 , 0\right)$ and $\left(3 , \frac{1}{9}\right)$.

$\left(3 , 0\right) \rightarrow$

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

$\left(3 , \frac{1}{9}\right) \rightarrow$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \left(3\right) {\left(\frac{1}{9}\right)}^{2}}{1 - 2 \left({3}^{2}\right) \left(\frac{1}{9}\right)} = \frac{\frac{2}{27}}{- 1} = - \frac{2}{27}$