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


Let's differentiate the equation \( 2x + 2 = \ln(y^2) \) and evaluate it at the point \((-1, 1)\).

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

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

1. **Differentiate the left-hand side**:
   \[
   \frac{d}{dx}(2x + 2) = 2
   \]

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

   Using the chain rule:
   \[
   \ln(y^2) = 2\ln(y)
   \]

   Differentiating \( 2\ln(y) \) with respect to \( x \) gives:
   \[
   \frac{d}{dx}[2\ln(y)] = 2 \cdot \frac{1}{y} \cdot \frac{dy}{dx} = \frac{2}{y} \cdot \frac{dy}{dx}
   \]

### Step 2: Set up the equation

Now, equate the derivatives:
\[
2 = \frac{2}{y} \cdot \frac{dy}{dx}
\]

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

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

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

Substitute \( y = 1 \):
\[
\frac{dy}{dx} = 1
\]

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

Would you like further clarification or details on any part of this process?

Here are 5 related questions you might find interesting:
1. How does the chain rule apply when differentiating logarithmic functions?
2. What are the implications of a derivative equaling 1 at a given point?
3. Can you compare this differentiation process to the one involving \( y^y \)?
4. How would the differentiation change if the logarithmic argument was more complex?
5. What role does implicit differentiation play in solving logarithmic equations?

**Tip:** Remember that logarithmic differentiation can simplify complex expressions, especially when dealing with powers or products inside the logarithm.

Learn how to differentiate the equation 2x + 2 = ln(y^2) using the chain rule and implicit differentiation. Find the derivative at the point (-1, 1) with step-by-step explanations. Suitable for advanced high school students and learners interested in calculus and logarithmic functions.

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