Find the derivative.
y equals 5 x squared e Superscript 3 x
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Part 1
A.StartFraction dy Over dx EndFraction equals 5 x e Superscript 3 x Baseline left parenthesis 3 x plus 2 right parenthesis nbsp
StartFraction dy Over dx EndFraction equals
5 x e Superscript 3 x Baseline left parenthesis 3 x plus 2 right parenthesis nbsp
B.StartFraction dy Over dx EndFraction equals 5 x e Superscript 3 x Baseline left parenthesis 2 x plus 3 right parenthesis nbsp
StartFraction dy Over dx EndFraction equals
5 x e Superscript 3 x Baseline left parenthesis 2 x plus 3 right parenthesis nbsp
C.StartFraction dy Over dx EndFraction equals 10 x e Superscript 3 x Baseline left parenthesis 2 x plus 3 right parenthesis
StartFraction dy Over dx EndFraction equals
10 x e Superscript 3 x Baseline left parenthesis 2 x plus 3 right parenthesis
D.StartFraction dy Over dx EndFraction equals 10 e x Superscript 3 x Baseline left parenthesis 3 x plus 2 right parenthesis nbsp
StartFraction dy Over dx EndFraction equals
10 e x Superscript 3 x Baseline left parenthesis 3 x plus 2 right parenthesis nbsp

To solve for the derivative of \( y = 5x^2 e^{3x} \), we can apply the **product rule** since this is a product of two functions: \( u(x) = 5x^2 \) and \( v(x) = e^{3x} \).

### Step-by-step solution:

1. **Product rule**: The derivative of a product \( u(x)v(x) \) is given by:
   \[
   \frac{d}{dx} \left( u(x) v(x) \right) = u'(x)v(x) + u(x)v'(x)
   \]

2. **Find the derivatives of \( u(x) = 5x^2 \)**:
   \[
   u'(x) = 10x
   \]

3. **Find the derivative of \( v(x) = e^{3x} \)** using the chain rule:
   \[
   v'(x) = e^{3x} \cdot 3 = 3e^{3x}
   \]

4. **Apply the product rule**:
   \[
   \frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)
   \]
   Substituting the values:
   \[
   \frac{dy}{dx} = 10x e^{3x} + 5x^2 \cdot 3e^{3x}
   \]
   Simplify:
   \[
   \frac{dy}{dx} = 10x e^{3x} + 15x^2 e^{3x}
   \]

5. **Factor out common terms**:
   \[
   \frac{dy}{dx} = 5x e^{3x} (2 + 3x)
   \]

Thus, the correct answer is:

\[
\frac{dy}{dx} = 5x e^{3x} (2 + 3x)
\]

This corresponds to **option B**.

Would you like more details on this? Here are some related questions you might find useful:

1. What is the product rule and how is it used in differentiation?
2. How does the chain rule work for functions involving exponentials?
3. Can you explain the significance of factoring in derivatives?
4. How do we handle more complicated products or quotients in differentiation?
5. What are some common mistakes to avoid when using the product rule?

**Tip:** Always check if you can simplify the result by factoring out common terms to make the expression more concise.

Find the derivative. y equals 5x^2 e^(3x).

Find the derivative of y = 5x^2 e^(3x) using the product rule and chain rule. This step-by-step solution explores how to apply these key calculus techniques, suitable for students in Grades 11-12.

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