To find the derivative of the given function:

$u = 3 \sin(r) \sqrt{\sin(r) + \sqrt{\sin(r)}}$

we'll apply both the product rule and chain rule. Let's go through it step-by-step.

1. Rewrite the expression for clarity. Note that we have: $u = 3 \sin(r) \left(\sin(r) + \sqrt{\sin(r)}\right)^{1/2}$

2. Differentiate using the product rule: Since $u$ is a product of two functions, $f(r) = 3 \sin(r)$ and $g(r) = \left(\sin(r) + \sqrt{\sin(r)}\right)^{1/2}$, we apply the product rule: $u' = f'(r)g(r) + f(r)g'(r)$

3. Find $f'(r)$: The derivative of $f(r) = 3 \sin(r)$ is: $f'(r) = 3 \cos(r)$

4. Differentiate $g(r)$: To find $g'(r)$, apply the chain rule to $g(r) = \left(\sin(r) + \sqrt{\sin(r)}\right)^{1/2}$: $g'(r) = \frac{1}{2} \left(\sin(r) + \sqrt{\sin(r)}\right)^{-1/2} \cdot \left(\cos(r) + \frac{\cos(r)}{2 \sqrt{\sin(r)}}\right)$

5. Combine terms to write the final derivative expression.

This results in:

$u' = 3 \cos(r) \sqrt{\sin(r) + \sqrt{\sin(r)}} + 3 \sin(r) \cdot \frac{\cos(r) + \frac{\cos(r)}{2 \sqrt{\sin(r)}}}{2 \sqrt{\sin(r) + \sqrt{\sin(r)}}}$

Would you like a more detailed step-by-step breakdown of the differentiation process or need help simplifying this derivative further?

Related Questions

1. How does the product rule work for derivatives in general?
2. Can you explain how the chain rule was applied in this example?
3. What is the role of the square root in this differentiation process?
4. How would the derivative change if the power on the expression were different?
5. Can we further simplify the expression for $u'$ here?

Tip

When differentiating products of functions that include square roots, always remember to apply the chain rule carefully within each square root term.