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.