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


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