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

$$\frac{dy}{dx}=\sec(\tan x)\tan(\tan x)\sec^2x$$
$$y=f(x)=\sec(\tan x)$$ Let $u=\tan x$, then $y=f(u)=\sec u$ We have $$\frac{dy}{du}=\frac{d}{du}(\sec u)=\sec u\tan u$$ $$\frac{du}{dx}=\frac{d}{dx}(\tan x)=\sec^2x$$ According to the Chain Rule: $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=\sec u\tan u\times\sec^2 x$$ Replace $u$ back with $\tan x$: $$\frac{dy}{dx}=\sec(\tan x)\tan(\tan x)\sec^2x$$