## Calculus 10th Edition

(a)$\frac{d}{dx}[sec(x)]=sec(x)tan(x)$ $\frac{d}{dx}[sec(x)]=\frac{d}{dx}[\frac{1}{cos(x)}]$ $= \frac{cos(x)(0)-(1)(-sin(x)}{cos(x)cos(x)} = \frac{sin(x)}{cos(x)cos(x)} = \frac{sin(x)}{cos(x)}\times\frac{1}{cos(x)} = tan(x)sec(x)$ (b))$\frac{d}{dx}[csc(x)]=-csc(x)cot(x)$ $\frac{d}{dx}[csc(x)]=\frac{d}{dx}[\frac{1}{sin(x)}]$ $\frac{sin(x)(0)-(1)(cos(x)}{sin(x)sin(x)}=\frac{-cos(x)}{sin(x)sin(x)}=\frac{-1}{sin(x)}\times\frac{cos(x)}{sin(x)}=-csc(x)cot(x)$ (c)$\frac{d}{dx}[cot(x)]=-csc^{2}(x)$ $\frac{d}{dx}[cot(x)]=\frac{d}{dx}[\frac{cos(x)}{sin(x)}]$ $=\frac{sin(x)[-sin(x)]-cos(x)cos(x)}{sin(x)sin(x)}=\frac{-sin^{2}(x)-cos^{2}(x)}{sin(x)sin(x)}=\frac{-[sin^{2}(x)+cos^{2}(x)]}{sin(x)sin(x)}=\frac{-1}{sin^{2}(x)}=-csc^{2}(x)$
Use Quotient Rule: $\frac{d}{dx}\frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)g(x)}$ (a)$\frac{d}{dx}[sec(x)]=sec(x)tan(x)$ $\frac{d}{dx}[sec(x)]=\frac{d}{dx}[\frac{1}{cos(x)}]$ $= \frac{cos(x)(0)-(1)(-sin(x)}{cos(x)cos(x)} = \frac{sin(x)}{cos(x)cos(x)} = \frac{sin(x)}{cos(x)}\times\frac{1}{cos(x)} = tan(x)sec(x)$ (b))$\frac{d}{dx}[csc(x)]=-csc(x)cot(x)$ $\frac{d}{dx}[csc(x)]=\frac{d}{dx}[\frac{1}{sin(x)}]$ $\frac{sin(x)(0)-(1)(cos(x)}{sin(x)sin(x)}=\frac{-cos(x)}{sin(x)sin(x)}=\frac{-1}{sin(x)}\times\frac{cos(x)}{sin(x)}=-csc(x)cot(x)$ (c)$\frac{d}{dx}[cot(x)]=-csc^{2}(x)$ $\frac{d}{dx}[cot(x)]=\frac{d}{dx}[\frac{cos(x)}{sin(x)}]$ $=\frac{sin(x)[-sin(x)]-cos(x)cos(x)}{sin(x)sin(x)}=\frac{-sin^{2}(x)-cos^{2}(x)}{sin(x)sin(x)}=\frac{-[sin^{2}(x)+cos^{2}(x)]}{sin(x)sin(x)}$ Use Trig Identity: $sin^{2}(x)+cos^{2}(x)=1$ $=\frac{-1}{sin^{2}(x)}=-csc^{2}(x)$