The expression given in the image is:

$e^{x^2 y^3} \left( 2xy^3 \ln(x^2) + 2x^{-1} \right)$

We are tasked with differentiating this expression with respect to $x$.

Step 1: Recognize the product rule

The expression is the product of two functions of $x$:

• The first function: $f(x) = e^{x^2 y^3}$
• The second function: $g(x) = \left( 2xy^3 \ln(x^2) + 2x^{-1} \right)$

Thus, we will apply the product rule: $\frac{d}{dx} \left( f(x) \cdot g(x) \right) = f'(x) \cdot g(x) + f(x) \cdot g'(x)$

Step 2: Differentiate $f(x) = e^{x^2 y^3}$

We use the chain rule here. The derivative of $e^{u}$ is $e^{u} \cdot u'$, where $u = x^2 y^3$: $f'(x) = e^{x^2 y^3} \cdot \frac{d}{dx}(x^2 y^3) = e^{x^2 y^3} \cdot (2x y^3)$

Step 3: Differentiate $g(x) = 2xy^3 \ln(x^2) + 2x^{-1}$

Now, let's differentiate the second part: $g'(x) = \frac{d}{dx} \left( 2xy^3 \ln(x^2) + 2x^{-1} \right)$

1. Differentiate $2xy^3 \ln(x^2)$: $\frac{d}{dx}(2xy^3 \ln(x^2)) = 2y^3 \left( \ln(x^2) + x \cdot \frac{d}{dx}(\ln(x^2)) \right)$ Since $\frac{d}{dx}(\ln(x^2)) = \frac{2}{x}$, this becomes: $2y^3 \left( \ln(x^2) + 1 \right)$

2. Differentiate $2x^{-1}$: $\frac{d}{dx}(2x^{-1}) = -2x^{-2}$

Thus, $g'(x) = 2y^3 \left( \ln(x^2) + 1 \right) - 2x^{-2}$

Step 4: Combine using the product rule

Now, apply the product rule: $\frac{d}{dx} \left( e^{x^2 y^3} \cdot \left( 2xy^3 \ln(x^2) + 2x^{-1} \right) \right)$ This becomes: $e^{x^2 y^3} \cdot (2x y^3) \cdot \left( 2xy^3 \ln(x^2) + 2x^{-1} \right) + e^{x^2 y^3} \cdot \left( 2y^3 \left( \ln(x^2) + 1 \right) - 2x^{-2} \right)$

You can expand and simplify further if needed, but this is the general result.

Would you like further clarification or a more detailed simplification?

Here are 5 related questions to expand on this concept:

1. How do you apply the product rule in more complex expressions?
2. What is the significance of the chain rule in differentiation?
3. How do logarithmic functions affect differentiation in product rule applications?
4. What are some common mistakes to avoid when differentiating products of exponential and logarithmic terms?
5. Can you explore other rules of differentiation such as the quotient rule?

Tip: When differentiating complex expressions, it often helps to break them into smaller parts and apply rules like the chain and product rules step by step.