To solve for $\frac{dy}{dx}$ given the equation:

$e^{2y} = xy^2 + \ln(x),$

we'll proceed by differentiating both sides with respect to $x$, applying implicit differentiation since $y$ is a function of $x$.

Step 1: Differentiate Both Sides with Respect to $x$

Left Side

The left side is $e^{2y}$. Differentiating with respect to $x$, we use the chain rule: $\frac{d}{dx} \left(e^{2y}\right) = e^{2y} \cdot \frac{d}{dx}(2y) = e^{2y} \cdot 2 \cdot \frac{dy}{dx}.$

Right Side

The right side is $xy^2 + \ln(x)$. Differentiating each term separately:

1. For $xy^2$, use the product rule: $\frac{d}{dx} (xy^2) = y^2 \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(y^2).$ This becomes: $y^2 + x \cdot 2y \cdot \frac{dy}{dx} = y^2 + 2xy \frac{dy}{dx}.$

2. For $\ln(x)$, the derivative is: $\frac{d}{dx} (\ln(x)) = \frac{1}{x}.$

Putting these together, the derivative of the right side is: $y^2 + 2xy \frac{dy}{dx} + \frac{1}{x}.$

Step 2: Set Up the Equation

Now, we equate the derivatives of both sides: $e^{2y} \cdot 2 \frac{dy}{dx} = y^2 + 2xy \frac{dy}{dx} + \frac{1}{x}.$

Step 3: Solve for $\frac{dy}{dx}$

Collect all terms involving $\frac{dy}{dx}$ on one side: $2e^{2y} \frac{dy}{dx} - 2xy \frac{dy}{dx} = y^2 + \frac{1}{x}.$

Factor out $\frac{dy}{dx}$: $\frac{dy}{dx} (2e^{2y} - 2xy) = y^2 + \frac{1}{x}.$

Finally, solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{y^2 + \frac{1}{x}}{2e^{2y} - 2xy}.$

Final Answer

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

Would you like further explanation on any step?

Related Questions

1. How does implicit differentiation work, and why is it useful?
2. What is the product rule in differentiation, and how is it applied here?
3. How does the chain rule apply when differentiating $e^{2y}$ with respect to $x$?
4. What are the key differences between implicit and explicit differentiation?
5. How would this approach change if we had $e^{3y}$ instead of $e^{2y}$?

Tip

Always remember to apply the chain rule carefully when differentiating terms with $y$ as a function of $x$.