To differentiate the function $f(\theta) = \frac{\sin(9\theta)}{\theta}$, we will use the quotient rule. The quotient rule states that if we have a function in the form $\frac{u}{v}$, then its derivative is:

$f'(\theta) = \frac{u'v - uv'}{v^2}$

Here:

• $u = \sin(9\theta)$
• $v = \theta$

We need to find $u'$ and $v'$:

1. Differentiate $u = \sin(9\theta)$: $u' = 9 \cos(9\theta)$

2. Differentiate $v = \theta$: $v' = 1$

Now, applying the quotient rule:

$f'(\theta) = \frac{(9 \cos(9\theta)) \cdot \theta - \sin(9\theta) \cdot 1}{\theta^2}$

Simplifying this expression:

$f'(\theta) = \frac{9 \theta \cos(9\theta) - \sin(9\theta)}{\theta^2}$

So, the derivative is:

$f'(\theta) = \frac{9 \theta \cos(9\theta) - \sin(9\theta)}{\theta^2}$

Would you like a further breakdown or explanation of each step?

Here are some related questions:

1. How would the result change if $f(\theta)$ had a different coefficient inside the sine function, like $\sin(7\theta)$?
2. Can the same differentiation technique be applied if $f(\theta)$ were a product instead of a quotient?
3. How does the differentiation process change if $\theta$ approaches zero?
4. What happens if we apply L'Hôpital's Rule to the function $f(\theta)$ as $\theta \to 0$?
5. How can trigonometric identities simplify differentiation of more complex trigonometric functions?

Tip: When using the quotient rule, always simplify the numerator and denominator separately to avoid errors.