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.