## Calculus (3rd Edition)

$$H'(t)=\sec t+2\sin t\sec^2 t\tan t$$
Recall the product rule: $(uv)'=u'v+uv'$ Recall that $(\sin x)'=\cos x$. Recall that $(\sec x)'=\sec x\tan x$. Since $H(t)=\sin t\sec^2t$, then using the product rule, we have \begin{align*} H'(t)&=(\sin t)'\sec^2t+\sin t(\sec^2 t)'\\ &=\cos t\sec^2t+2(\sin t) (\sec t) \sec t \tan t\\ &=\cos t\sec^2t+2\sin t\sec^2 t\tan t\\ &=\sec t+2\sin t\sec^2 t\tan t. \end{align*} (Where we used the fact that $\sec t=\dfrac{1}{\cos t}$ in the last line.)