## Calculus: Early Transcendentals 8th Edition

$$\frac{df}{dt}=-\sec^2(\sec(\cos t))\times\sec(\cos t)\times\tan(\cos t)\times(\sin t)$$
$$f(t)=\tan(\sec(\cos t))$$ Then, $$\frac{df}{dt}=\frac{d(\tan(\sec(\cos t)))}{dt}$$ Let $u=\cos t$, then $f=\tan(\sec u)$. According to the Chain Rule, we have $$\frac{df}{dt}=\frac{df}{du}\frac{du}{dt}$$ $$\frac{df}{dt}=\frac{d(\tan(\sec u))}{du}\frac{d(\cos t)}{dt}$$ $$\frac{df}{dt}=\frac{dg}{du}(-\sin t)$$ Doing the Chain Rule one more time for $\frac{dg}{du}$. Let $v=\sec u$, then $g=\tan v$. Applying the Chain Rule, we have $$\frac{dg}{du}=\frac{dg}{dv}\frac{dv}{du}$$ $$\frac{dg}{du}=\frac{d(\tan v)}{dv}\frac{d(\sec u)}{du}$$ $$\frac{dg}{du}=\sec^2 v\times(\sec u\tan u)$$ $$\frac{dg}{du}=\sec^2(\sec u)\times\sec u\times\tan u$$ Therefore, $$\frac{df}{dt}=\sec^2(\sec u)\times\sec u\times\tan u\times(-\sin t)$$ $$\frac{df}{dt}=-\sec^2(\sec(\cos t))\times\sec(\cos t)\times\tan(\cos t)\times(\sin t)$$