also differentiate for 2x+2=ln(yy) at (-1,1)

To differentiate the equation \( 2x + 2 = \ln(y^y) \) and evaluate it at the point \((-1, 1)\), we'll follow these steps:

### Step 1: Differentiate both sides with respect to \( x \)

Given:
\[
2x + 2 = \ln(y^y)
\]

1. **Differentiate the left-hand side**:

\[
\frac{d}{dx}(2x + 2) = 2
\]

2. **Differentiate the right-hand side**:
   \[
   \frac{d}{dx}[\ln(y^y)]
   \]

   Using the chain rule and the fact that \( y^y = e^{y \ln(y)} \), we get:
   \[
   \ln(y^y) = y \ln(y)
   \]
   
   Differentiating with respect to \( x \), we use the product rule:
   \[
   \frac{d}{dx}[y \ln(y)] = \frac{dy}{dx} \ln(y) + \frac{dy}{dx} + y \cdot \frac{1}{y} \cdot \frac{dy}{dx}
   \]
   
   Simplifying, this becomes:
   \[
   \frac{dy}{dx} \cdot (\ln(y) + 1)
   \]

### Step 2: Set up the equation

Now, equate the derivatives:
\[
2 = \frac{dy}{dx} \cdot (\ln(y) + 1)
\]

### Step 3: Solve for \(\frac{dy}{dx}\)

To solve for \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = \frac{2}{\ln(y) + 1}
\]

### Step 4: Evaluate at the point \((-1, 1)\)

Substitute \( y = 1 \):
\[
\frac{dy}{dx} = \frac{2}{\ln(1) + 1}
\]
Since \(\ln(1) = 0\):
\[
\frac{dy}{dx} = \frac{2}{0 + 1} = 2
\]

So, the derivative at the point \((-1, 1)\) is \( \frac{dy}{dx} = 2 \).

Would you like any additional details or explanations on this process? 

Here are 5 related questions you might find interesting:
1. How do you differentiate an equation involving a product of functions?
2. What is the significance of evaluating a derivative at a specific point?
3. How does the chain rule apply to functions within logarithms?
4. Can you apply implicit differentiation to more complex equations?
5. What are the differences between explicit and implicit differentiation?

**Tip:** Always remember that logarithms of 1 are zero, which can simplify calculations in differentiation.

Learn how to differentiate the equation 2x + 2 = ln(yy) with respect to x and evaluate the derivative at the point (-1, 1). This problem involves the chain rule and logarithmic differentiation, providing a detailed step-by-step solution.

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