Questions: Differentiate f(t) = sec t / (-3 + sec t).

Solution Steps

To differentiate the function \( f(t) = \frac{\sec t}{-3 + \sec t} \), we will use the quotient rule for differentiation. The quotient rule states that if you have a function \( \frac{u(t)}{v(t)} \), its derivative is given by \( \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2} \). Here, \( u(t) = \sec t \) and \( v(t) = -3 + \sec t \).

1. Identify \( u(t) \) and \( v(t) \).
2. Compute \( u'(t) \) and \( v'(t) \).
3. Apply the quotient rule to find \( f'(t) \).

We define the functions as follows: \[ u(t) = \sec t = \frac{1}{\cos t} \] \[ v(t) = -3 + \sec t = -3 + \frac{1}{\cos t} \]

Next, we differentiate \( u(t) \) and \( v(t) \): \[ u'(t) = \frac{d}{dt} \left( \sec t \right) = \sec t \tan t = \frac{\sin t}{\cos^2 t} \] \[ v'(t) = \frac{d}{dt} \left( -3 + \sec t \right) = \sec t \tan t = \frac{\sin t}{\cos^2 t} \]

Using the quotient rule, we find the derivative \( f'(t) \): \[ f'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2} \] Substituting the values we computed: \[ f'(t) = \frac{\left(-3 + \frac{1}{\cos t}\right) \frac{\sin t}{\cos^2 t} - \frac{\sin t}{\cos^2 t} \left(-3 + \frac{1}{\cos t}\right)}{\left(-3 + \frac{1}{\cos t}\right)^2} \] This simplifies to: \[ f'(t) = \frac{-3 \sin t}{(3 \cos t - 1)^2} \]

Final Answer