Untitled

To differentiate the expression \( \frac{5 + 3\cos(\theta)}{6\sin(\theta)} \) with respect to \(\theta\), we will use the quotient rule. The quotient rule for differentiation states that if we have a function \( f(\theta) = \frac{u(\theta)}{v(\theta)} \), then its derivative is given by:

\[
f'(\theta) = \frac{u'(\theta)v(\theta) - u(\theta)v'(\theta)}{[v(\theta)]^2}
\]

### Step-by-Step Differentiation

1. **Identify \( u(\theta) \) and \( v(\theta) \):**
   - \( u(\theta) = 5 + 3\cos(\theta) \)
   - \( v(\theta) = 6\sin(\theta) \)

2. **Differentiate \( u(\theta) \) and \( v(\theta) \):**
   - \( u'(\theta) = \frac{d}{d\theta} [5 + 3\cos(\theta)] \)
     \[
     u'(\theta) = 0 + 3(-\sin(\theta)) = -3\sin(\theta)
     \]

   - \( v'(\theta) = \frac{d}{d\theta} [6\sin(\theta)] \)
     \[
     v'(\theta) = 6\cos(\theta)
     \]

3. **Apply the quotient rule:**
   \[
   f'(\theta) = \frac{u'(\theta)v(\theta) - u(\theta)v'(\theta)}{[v(\theta)]^2}
   \]
   Substitute \( u(\theta) \), \( v(\theta) \), \( u'(\theta) \), and \( v'(\theta) \):
   \[
   f'(\theta) = \frac{(-3\sin(\theta))(6\sin(\theta)) - (5 + 3\cos(\theta))(6\cos(\theta))}{[6\sin(\theta)]^2}
   \]

4. **Simplify the numerator:**
   \[
   \text{Numerator} = (-3\sin(\theta))(6\sin(\theta)) - (5 + 3\cos(\theta))(6\cos(\theta))
   \]
   \[
   = -18\sin^2(\theta) - (30\cos(\theta) + 18\cos^2(\theta))
   \]
   \[
   = -18\sin^2(\theta) - 30\cos(\theta) - 18\cos^2(\theta)
   \]

5. **Simplify the denominator:**
   \[
   \text{Denominator} = [6\sin(\theta)]^2 = 36\sin^2(\theta)
   \]

6. **Combine the numerator and denominator:**
   \[
   f'(\theta) = \frac{-18\sin^2(\theta) - 30\cos(\theta) - 18\cos^2(\theta)}{36\sin^2(\theta)}
   \]

7. **Factor and simplify if possible:**
   \[
   f'(\theta) = \frac{-18(\sin^2(\theta) + \cos^2(\theta)) - 30\cos(\theta)}{36\sin^2(\theta)}
   \]
   Using the Pythagorean identity \(\sin^2(\theta) + \cos^2(\theta) = 1\):
   \[
   f'(\theta) = \frac{-18(1) - 30\cos(\theta)}{36\sin^2(\theta)}
   \]
   \[
   = \frac{-18 - 30\cos(\theta)}{36\sin^2(\theta)}
   \]
   Simplify the fraction:
   \[
   = \frac{-18 - 30\cos(\theta)}{36\sin^2(\theta)} = \frac{-18(1 + \frac{5\cos(\theta)}{3})}{36\sin^2(\theta)}
   \]
   \[
   = \frac{-18(3 + 5\cos(\theta))}{108\sin^2(\theta)} = \frac{-3(3 + 5\cos(\theta))}{18\sin^2(\theta)}
   \]
   \[
   = \frac{- (3 + 5\cos(\theta))}{6\sin^2(\theta)}
   \]

So, the derivative of \( \frac{5 + 3\cos(\theta)}{6\sin(\theta)} \) with respect to \(\theta\) is:

\[
f'(\theta) = \frac{-(3 + 5\cos(\theta))}{6\sin^2(\theta)}
\]