## Calculus: Early Transcendentals 8th Edition

1) Rewrite $\sec x$ into $\frac{1}{cos x}$ 2) Then apply the Quotient Rule for $\frac{d}{dx}(\frac{1}{cos x})$ 3) Use trigonometrical rules to simplify.
We know that $\sec x=\frac{1}{\cos x}$. So, $$\frac{d}{dx}(\sec x)=\frac{d}{dx}\Bigg(\frac{1}{\cos x}\Bigg)$$ Now apply the Quotient Rule, we have $$\frac{d}{dx}(\sec x)=\frac{(1)'\cos x-1(\cos x)'}{(\cos x)^2}$$ $$\frac{d}{dx}(\sec x)=\frac{0\times\cos x-(-\sin x)}{\cos^2 x}$$ $$\frac{d}{dx}(\sec x)=\frac{\sin x}{\cos^2 x}$$ $$\frac{d}{dx}(\sec x)=\frac{\sin x}{\cos x}\times\frac{1}{\cos x}$$ $$\frac{d}{dx}(\sec x)=\tan x\sec x$$ The statement has been proved.