https://file.aiquickdraw.com/mathbot/file/1727943490549kbpskhpw.jpg

The expression given in the image is:

\[
2xy^3 e^{x^2 y^3} \cdot \ln(x^2) + e^{x^2 y^3} \cdot \frac{2}{x}
\]

Let's differentiate this with respect to \(x\). We will use the product rule, chain rule, and other relevant differentiation techniques.

### Step-by-step differentiation:

1. **Differentiate the first term: \( 2xy^3 e^{x^2 y^3} \ln(x^2) \):**
   - Use the product rule: \( (f \cdot g)' = f' \cdot g + f \cdot g' \).
   - Let \( f = 2xy^3 e^{x^2 y^3} \) and \( g = \ln(x^2) \).

   - **Differentiate \( f = 2xy^3 e^{x^2 y^3} \):**
     - Differentiate \(2xy^3\) using the product rule: \( 2 \cdot (xy^3) \).
     - Derivative of \(x\) is 1, so the derivative of \(2xy^3\) is \(2y^3\).
     - Now, apply the chain rule for \(e^{x^2 y^3}\). The derivative of \(e^u\) is \(e^u \cdot u'\), where \(u = x^2 y^3\).
       - Derivative of \(x^2 y^3\) with respect to \(x\) is \(2xy^3\).
     - Therefore, the derivative of \( 2xy^3 e^{x^2 y^3} \) is:
       \[
       2y^3 e^{x^2 y^3} + 2xy^3 e^{x^2 y^3} \cdot 2xy^3 = 2y^3 e^{x^2 y^3} + 4x^2 y^6 e^{x^2 y^3}
       \]

   - **Differentiate \(g = \ln(x^2)\):**
     - The derivative of \( \ln(x^2) \) is \( \frac{2}{x} \).

   Now, using the product rule:
   \[
   (f \cdot g)' = \left( 2y^3 e^{x^2 y^3} + 4x^2 y^6 e^{x^2 y^3} \right) \ln(x^2) + 2xy^3 e^{x^2 y^3} \cdot \frac{2}{x}
   \]

2. **Differentiate the second term: \( e^{x^2 y^3} \cdot \frac{2}{x} \):**
   - Use the product rule again.
   - Let \( f = e^{x^2 y^3} \) and \( g = \frac{2}{x} \).

   - **Differentiate \( f = e^{x^2 y^3} \):**
     - As before, the derivative of \(e^{x^2 y^3}\) is \( e^{x^2 y^3} \cdot 2xy^3 \).
   - **Differentiate \(g = \frac{2}{x}\):**
     - The derivative of \( \frac{2}{x} \) is \( -\frac{2}{x^2} \).

   Now, using the product rule:
   \[
   \left( e^{x^2 y^3} \cdot \frac{2}{x} \right)' = 2xy^3 e^{x^2 y^3} \cdot \frac{2}{x} + e^{x^2 y^3} \cdot \left( -\frac{2}{x^2} \right)
   \]

### Final expression:

After applying the product and chain rules to both terms, the derivative of the full expression is:

\[
\left( 2xy^3 e^{x^2 y^3} \ln(x^2) + e^{x^2 y^3} \cdot \frac{2}{x} \right)' = \left( 2y^3 e^{x^2 y^3} + 4x^2 y^6 e^{x^2 y^3} \right) \ln(x^2) + 2xy^3 e^{x^2 y^3} \cdot \frac{2}{x} + 2xy^3 e^{x^2 y^3} \cdot \frac{2}{x} + e^{x^2 y^3} \cdot \left( -\frac{2}{x^2} \right)
\]

Let me know if you'd like more details or clarifications.

### Related questions:
1. How does the chain rule work for exponential functions like \( e^{x^2 y^3} \)?
2. What is the product rule, and how is it applied in differentiation?
3. Why do we need to use the logarithmic differentiation for \( \ln(x^2) \)?
4. How would the derivative change if \(y\) was a function of \(x\) instead of a constant?
5. What is the significance of simplifying expressions after differentiation?

### Tip:
Always apply the product rule whenever you're differentiating a product of two or more functions to avoid mistakes.

Differentiate with respect to x: 2xy^3 * e^(x^2y^3) * ln(x^2) + e^(x^2y^3) * (2/x)

Learn how to differentiate the expression 2xy^3 * e^(x^2y^3) * ln(x^2) + e^(x^2y^3) * (2/x) with detailed step-by-step solutions. This problem uses key calculus concepts such as the product rule, chain rule, and logarithmic differentiation. Suitable for high school calculus students.

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