Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 49

Find the first and second derivatives of $H(t) = \tan 3t$
Solving for the first derivative of the given function


$
\begin{equation}
\begin{aligned}
H'(t) &= \frac{d}{dt} ( \tan 3t)\\
\\
H'(t) &= \sec^2 3t \cdot \frac{d}{dt} (3t)\\
\\
H'(t) &= (\sec^2 3t) (3)(1)\\
\\
H'(t) &= 3\sec^2 3t
\end{aligned}
\end{equation}
$



Solving for the second derivative of the given function


$
\begin{equation}
\begin{aligned}
H''(t) &= \frac{d}{dt} (3\sec^2t)\\
\\
H''(t) &= 3 \cdot \frac{d}{dt} ( 3\sec3t)^2\\
\\
H''(t) &= (3)(2)(\sec 3t) \cdot \frac{d}{dt} (\sec 3t)\\
\\
H''(t) &= 6 \sec 3t \cdot \sec 3t \tan 3t \cdot \frac{d}{dt}(3t)\\
\\
H''(t) &= 6 \sec 3t \cdot \sec 3t \tan 3t \cdot 3\\
\\
H''(t) &= 18 \sec^2(3t)\tan(3t)
\end{aligned}
\end{equation}
$