Question

Find the derivative of the function.
f(t) =
6

1 + tan t

Answer 1

To find the derivative of the function \( f(t) = \frac{6}{1 + \tan t} \), we can use the quotient rule. The quotient rule states that if you have a function in the form of \( \frac{u}{v} \), then the derivative is given by:

\[
f'(t) = \frac{u'v - uv'}{v^2}
\]

For our function, we can identify:
- \( u = 6 \)
- \( v = 1 + \tan t \)

Now we need to find the derivatives \( u' \) and \( v' \):
- \( u' = 0 \) (since the derivative of a constant is zero)
- \( v' = \sec^2 t \) (the derivative of \( \tan t \) is \( \sec^2 t \))

Now we can apply the quotient rule:

\[
f'(t) = \frac{0 \cdot (1 + \tan t) - 6 \cdot \sec^2 t}{(1 + \tan t)^2}
\]

This simplifies to:

\[
f'(t) = \frac{-6 \sec^2 t}{(1 + \tan t)^2}
\]

Therefore, the derivative of the function \( f(t) \) is:

\[
f'(t) = \frac{-6 \sec^2 t}{(1 + \tan t)^2}
\]