Let's solve the problem using implicit differentiation.

The given equation is:

$x \ln(y) + 5y = 2x^3$

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

We will use the product rule, chain rule, and implicit differentiation.

Left side:

1. $\frac{d}{dx} \left( x \ln(y) \right) = \ln(y) + x \cdot \frac{1}{y} \cdot y'$ (product rule and chain rule for $\ln(y)$).

2. $\frac{d}{dx} \left( 5y \right) = 5y'$.

Right side:

1. $\frac{d}{dx} \left( 2x^3 \right) = 6x^2$.

Thus, after differentiation: $\ln(y) + \frac{x}{y} \cdot y' + 5y' = 6x^2$

Step 2: Solve for $y'$.

Rearranging the terms to isolate $y'$:

$\frac{x}{y} \cdot y' + 5y' = 6x^2 - \ln(y)$

Factor out $y'$:

$y' \left( \frac{x}{y} + 5 \right) = 6x^2 - \ln(y)$

Now, solve for $y'$:

$y' = \frac{6x^2 - \ln(y)}{\frac{x}{y} + 5}$

Step 3: Evaluate $y'$ at $(-1, 1)$.

Substitute $x = -1$ and $y = 1$:

$y' = \frac{6(-1)^2 - \ln(1)}{\frac{-1}{1} + 5}$

Since $\ln(1) = 0$, the equation simplifies to:

$y' = \frac{6(1) - 0}{-1 + 5} = \frac{6}{4} = \frac{3}{2}$

Final Answers:

• $y' = \frac{6x^2 - \ln(y)}{\frac{x}{y} + 5}$
• $y'(-1, 1) = \frac{3}{2}$

Would you like more details or have any questions?

Here are 5 related questions for further practice:

1. What is implicit differentiation, and how does it differ from explicit differentiation?
2. How does the chain rule apply when differentiating $\ln(y)$?
3. What is the product rule, and how is it applied in this problem?
4. How would the solution change if the equation involved $\ln(x)$ instead of $\ln(y)$?
5. How can implicit differentiation be used in solving related rates problems?

Tip: Always simplify logarithmic terms like $\ln(1) = 0$ early to make the differentiation process easier.