y(x)=e^0.61xsin(−1.04x) find the derivitve

To find the derivative of the function \( y(x) = e^{0.61x} \sin(-1.04x) \), we will use the product rule and the chain rule.

### Step-by-Step Solution:

Given:
\[
y(x) = e^{0.61x} \sin(-1.04x)
\]

1. **Identify the functions to differentiate:**
   - Let \( u(x) = e^{0.61x} \)
   - Let \( v(x) = \sin(-1.04x) \)

2. **Differentiate \( u(x) \):**
   \[
   u'(x) = \frac{d}{dx} \left( e^{0.61x} \right) = 0.61 e^{0.61x}
   \]

3. **Differentiate \( v(x) \):**
   \[
   v'(x) = \frac{d}{dx} \left( \sin(-1.04x) \right) = -1.04 \cos(-1.04x)
   \]
   (Note: The derivative of \( \sin(ax) \) is \( a \cos(ax) \), and \(\cos(-\theta) = \cos(\theta)\).)

4. **Apply the product rule:**
   The product rule states that if \( y(x) = u(x) \cdot v(x) \), then:
   \[
   y'(x) = u'(x) v(x) + u(x) v'(x)
   \]
   Substituting the values:
   \[
   y'(x) = 0.61 e^{0.61x} \sin(-1.04x) + e^{0.61x} \left(-1.04 \cos(-1.04x)\right)
   \]

5. **Simplify the expression:**
   \[
   y'(x) = e^{0.61x} \left( 0.61 \sin(-1.04x) - 1.04 \cos(-1.04x) \right)
   \]

### Final Answer:
\[
y'(x) = e^{0.61x} \left( 0.61 \sin(-1.04x) - 1.04 \cos(-1.04x) \right)
\]

Would you like more details on any part of this solution?

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Here are 5 related questions you might find interesting:

1. How do you apply the chain rule to composite functions?
2. What are the basic rules for differentiating trigonometric functions?
3. How do you differentiate a product of exponential and trigonometric functions?
4. Can the product rule be extended to more than two functions?
5. What is the derivative of \( y(x) = e^{kx} \cos(ax) \) for arbitrary constants \( k \) and \( a \)?

**Tip:** Always check the argument of trigonometric functions before differentiating to avoid sign mistakes!

Learn how to find the derivative of the function y(x) = e^0.61x sin(-1.04x) using the product rule and chain rule. This detailed step-by-step solution covers differentiation of exponential and trigonometric functions. Suitable for advanced high school and college students.

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