## University Calculus: Early Transcendentals (3rd Edition)

$$\frac{dy}{dt}=10(t\tan t)^9\Big(\tan t+t\sec^2t\Big)$$
$$\frac{dy}{dt}=\frac{d}{dt}(t\tan t)^{10}$$ According to the Chain Rule: $$\frac{dy}{dt}=10(t\tan t)^9\frac{d}{dt}(t\tan t)=10(t\tan t)^9\Big(\tan t+t(\tan t)'\Big)$$ $$\frac{dy}{dt}=10(t\tan t)^9\Big(\tan t+t\sec^2t\Big)$$