Question

1. - / 5.55 points

differentiate.
$f(\theta) = \frac{\sec \theta}{4 + \sec \theta}$
$f(\theta) = \square$

Explanation:

Step1: Apply quotient rule

Let \( u = \sec\theta \), \( v = 4 + \sec\theta \). Quotient rule: \( f'(\theta) = \frac{u'v - uv'}{v^2} \)

Step2: Compute derivatives of u and v

\( u' = \sec\theta\tan\theta \), \( v' = \sec\theta\tan\theta \)

Step3: Substitute into quotient rule

\( f'(\theta) = \frac{(\sec\theta\tan\theta)(4+\sec\theta) - (\sec\theta)(\sec\theta\tan\theta)}{(4+\sec\theta)^2} \)

Step4: Simplify numerator

\( = \frac{4\sec\theta\tan\theta + \sec^2\theta\tan\theta - \sec^2\theta\tan\theta}{(4+\sec\theta)^2} = \frac{4\sec\theta\tan\theta}{(4+\sec\theta)^2} \)

Answer:

\( \frac{4\sec\theta\tan\theta}{(4 + \sec\theta)^2} \)