We are given the implicit equation:

$y + x \cos(y) = x^2 y$

We need to find $\frac{dy}{dx}$, which is the derivative of $y$ with respect to $x$.

Step 1: Differentiate both sides with respect to $x$

Left side:

1. The derivative of $y$ with respect to $x$ is $\frac{dy}{dx}$ (using the chain rule).
2. For $x \cos(y)$, we apply the product rule. The derivative of $x$ is 1, and the derivative of $\cos(y)$ with respect to $x$ is $-\sin(y) \cdot \frac{dy}{dx}$ (again using the chain rule).

So, the derivative of the left-hand side is: $\frac{d}{dx}(y) + \frac{d}{dx}(x \cos(y)) = \frac{dy}{dx} + \left( \cos(y) - x \sin(y) \frac{dy}{dx} \right)$

Right side:

For the term $x^2 y$, we apply the product rule: $\frac{d}{dx}(x^2 y) = 2xy + x^2 \frac{dy}{dx}$

Step 2: Combine everything

Now we set the derivatives equal to each other:

$\frac{dy}{dx} + \left( \cos(y) - x \sin(y) \frac{dy}{dx} \right) = 2xy + x^2 \frac{dy}{dx}$

Step 3: Collect all terms with $\frac{dy}{dx}$

Group the terms involving $\frac{dy}{dx}$ on one side:

$\frac{dy}{dx} - x \sin(y) \frac{dy}{dx} - x^2 \frac{dy}{dx} = 2xy - \cos(y)$

Factor out $\frac{dy}{dx}$ on the left-hand side:

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

Step 4: Solve for $\frac{dy}{dx}$

Finally, solve for $\frac{dy}{dx}$:

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

Final Answer:

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

Would you like to go over any specific part of the solution in more detail? Or do you have any further questions?

Related questions:

1. How do you handle higher-order derivatives when solving implicit equations?
2. Can you solve this for $x$ in terms of $y$?
3. How do you find the second derivative $y''$ of an implicit function?
4. How does the chain rule apply when differentiating trigonometric functions with $y$ as a variable?
5. What is the significance of implicit differentiation in applications like physics?

Tip: When differentiating implicitly, always remember to treat $y$ as a function of $x$, so whenever you differentiate terms involving $y$, use the chain rule.