The equation given for implicit differentiation is:

$\ln(x^2 + 5y) = \sin(xy)$

To find the derivative $\frac{dy}{dx}$, we will differentiate both sides of the equation with respect to $x$, applying the chain rule where necessary.

Step-by-step solution:

Left-hand side:

$\frac{d}{dx} \left( \ln(x^2 + 5y) \right)$ Using the chain rule, this becomes: $\frac{1}{x^2 + 5y} \cdot \left( 2x + 5 \frac{dy}{dx} \right)$ Here, $2x$ is the derivative of $x^2$, and $5 \frac{dy}{dx}$ is the derivative of $5y$ with respect to $x$, where we applied the chain rule to differentiate $y$ implicitly.

Right-hand side:

$\frac{d}{dx} \left( \sin(xy) \right)$ Using the product and chain rules, we get: $\cos(xy) \cdot \left( y + x \frac{dy}{dx} \right)$ Here, $y$ is the derivative of $xy$ with respect to $x$, and $x \frac{dy}{dx}$ is the implicit differentiation of $y$.

Combining both sides:

Now, we have the equation: $\frac{1}{x^2 + 5y} \cdot \left( 2x + 5 \frac{dy}{dx} \right) = \cos(xy) \cdot \left( y + x \frac{dy}{dx} \right)$

Solving for $\frac{dy}{dx}$:

To isolate $\frac{dy}{dx}$, expand both sides and group all terms containing $\frac{dy}{dx}$ on one side.

Would you like to continue with the full solution, or any specific part of this step detailed further?

Here are some questions related to this process:

1. Why do we apply the chain rule when differentiating $\ln(x^2 + 5y)$?
2. What is the role of implicit differentiation in handling $y$ in the equation?
3. How does the product rule apply to the term $\sin(xy)$?
4. Could you solve for $\frac{dy}{dx}$ explicitly by expanding both sides of the equation?
5. What would change in the process if the equation involved $\cos(x^2 + y^2)$?

Tip: When performing implicit differentiation, always treat $y$ as a function of $x$ and apply the chain rule when differentiating terms involving $y$.