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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.

Use implicit differentiation to find y' and evaluate y' at (-1,1) for the equation x ln(y) + 5y = 2x^3.

Learn how to solve the equation x ln(y) + 5y = 2x^3 using implicit differentiation. This problem involves key calculus concepts such as the chain rule, product rule, and logarithmic differentiation. The solution is evaluated at the point (-1,1). Suitable for students in Grades 11-12.

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