se the product rule to solve g(x)=**√** xSinx

To differentiate \( g(x) = \sqrt{x} \sin(x) \) using the **product rule**, let’s follow these steps carefully.

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### **Product Rule:**

If \( g(x) = u(x) \cdot v(x) \), then:
\[
g'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)
\]

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### **Step-by-Step Solution:**

In our case:
- \( u(x) = \sqrt{x} = x^{1/2} \)
- \( v(x) = \sin(x) \)

Now, let’s find the derivatives of \( u(x) \) and \( v(x) \) individually.

1. **Derivative of \( u(x) = x^{1/2} \)**:
   \[
   u'(x) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}
   \]

2. **Derivative of \( v(x) = \sin(x) \)**:
   \[
   v'(x) = \cos(x)
   \]

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### **Apply the Product Rule:**

\[
g'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)
\]

Substitute \( u'(x) = \frac{1}{2\sqrt{x}} \), \( u(x) = \sqrt{x} \), \( v(x) = \sin(x) \), and \( v'(x) = \cos(x) \):

\[
g'(x) = \frac{1}{2\sqrt{x}} \cdot \sin(x) + \sqrt{x} \cdot \cos(x)
\]

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### **Final Answer:**

\[
g'(x) = \frac{\sin(x)}{2\sqrt{x}} + \sqrt{x} \cos(x)
\]

This is the derivative of \( g(x) = \sqrt{x} \sin(x) \) using the product rule.

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Do you want further clarification or additional steps?

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### **5 Related Questions:**
1. How do you apply the chain rule in derivatives?
2. How can you simplify \( \frac{\sin(x)}{2\sqrt{x}} + \sqrt{x} \cos(x) \) further?
3. What happens if we differentiate the answer again (second derivative)?
4. How does the product rule work with three functions?
5. What are some real-world applications of the product rule?

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**Tip:** When working with square roots, it’s often useful to rewrite them as fractional exponents to simplify differentiation.

Use the product rule to solve g(x)=√x * sin(x)

Learn to differentiate g(x) = √x * sin(x) by applying the product rule. This solution covers key calculus concepts, including differentiation, power rule, and the product rule for derivatives, with step-by-step guidance. Ideal for students in Grades 11-12.

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