Find s' with respect to t for s = 1/4 tan^4t - 1/2 tan²t

Find $s'$ with respect to $t$ for $s = \dfrac{1}{4}\tan^{4}t - \dfrac{1}{2}\tan^{2}t - \ln\cos t$.

Step 1: Differentiate each term

Writing $\sec^{2}t = \dfrac{1}{\cos^{2}t}$ for the derivative of the tangent, and using Problem 120 for the last term: \[\frac{d}{dt}\!\left(\tfrac14\tan^{4}t\right) = \frac{\tan^{3}t}{\cos^{2}t}, \qquad \frac{d}{dt}\!\left(\tfrac12\tan^{2}t\right) = \frac{\tan t}{\cos^{2}t}, \qquad \frac{d}{dt}\bigl(-\ln\cos t\bigr) = \tan t.\]

Step 2: Combine and factor

\[ s' = \frac{\tan^{3}t}{\cos^{2}t} - \frac{\tan t}{\cos^{2}t} + \tan t = \tan t\!\left(\frac{\tan^{2}t-1}{\cos^{2}t}+1\right). \]

Step 3: Simplify the bracket

Substituting $\dfrac{1}{\cos^{2}t} = 1+\tan^{2}t$, the bracket becomes a difference of squares plus one: $\bigl(\tan^{2}t-1\bigr)\bigl(1+\tan^{2}t\bigr) + 1 = \tan^{4}t - 1 + 1 = \tan^{4}t$.

Step 4: Verification

\[t = \frac{\pi}{4}:\quad \tan t = 1 \;\Longrightarrow\; s' = 1; \qquad \frac{s(\pi/4+10^{-5})-s(\pi/4-10^{-5})}{2\times 10^{-5}} \approx 1.0 \;\;\checkmark\]

\[t = 0:\quad s' = 0, \text{ and } s(0) = 0 \;\;\checkmark\]