How do you implicitly differentiate #xy^2 + 2xy = 11-1/y#?

Question

Emma Aldridge · Accepted Answer

#y^2 + 2xy dy/dx + 2y + 2xdy/dx = 1/y^2 dy/dx#

Samantha Adkison · Answer

undefined To implicitly differentiate the equation $ xy^2 + 2xy = 11 - \frac{1}{y} $, follow these steps:

1. Differentiate each term with respect to $ x $.
2. Apply the product rule and chain rule where necessary.
3. Solve for $ \frac{{dy}}{{dx}} $ to find the derivative.

Here are the steps:

1. Differentiate $ xy^2 $ with respect to $ x $ using the product rule:
   $$ \frac{{d}}{{dx}}(xy^2) = y^2 \frac{{dx}}{{dx}} + x \frac{{d}}{{dx}}(y^2) $$
2. Differentiate $ 2xy $ with respect to $ x $:
   $$ \frac{{d}}{{dx}}(2xy) = 2y \frac{{dx}}{{dx}} + x \frac{{d}}{{dx}}(2y) $$
3. Differentiate $ 11 $ with respect to $ x $ (constant term):
   $$ \frac{{d}}{{dx}}(11) = 0 $$
4. Differentiate $ \frac{1}{y} $ with respect to $ x $ using the chain rule:
   $$ \frac{{d}}{{dx}}\left(\frac{1}{y}\right) = -\frac{1}{{y^2}} \frac{{dy}}{{dx}} $$

Now, plug these results back into the original equation and solve for $ \frac{{dy}}{{dx}} $:

$$ y^2 + 2xy + x \cdot 2y\frac{{dy}}{{dx}} = 0 - \frac{1}{{y^2}} \frac{{dy}}{{dx}} $$

Rearrange terms and solve for $ \frac{{dy}}{{dx}} $:

$$ 2xy\frac{{dy}}{{dx}} + 2y\frac{{dy}}{{dx}} + \frac{{1}}{{y^2}}\frac{{dy}}{{dx}} = -y^2 - 2xy $$

$$ \frac{{dy}}{{dx}}(2xy + 2y + \frac{{1}}{{y^2}}) = -y^2 - 2xy $$

$$ \frac{{dy}}{{dx}} = \frac{{-y^2 - 2xy}}{{2xy + 2y + \frac{{1}}{{y^2}}}} $$

That's the implicit derivative of the given equation.