在以下公式中，角度 *u* 假设以弧度表示。
d d x sin u = cos u {\displaystyle {\operatorname {d} \over \operatorname {d} x}\sin u=\cos u}
d d x cos u = − sin u {\displaystyle {\operatorname {d} \over \operatorname {d} x}\cos u=-\sin u}
d d x tan u = sec 2 u {\displaystyle {\operatorname {d} \over \operatorname {d} x}\tan u=\sec ^{2}u}
d d x cot u = − csc 2 u {\displaystyle {\operatorname {d} \over \operatorname {d} x}\cot u=-\csc ^{2}u}
d d x sec u = sec u tan u {\displaystyle {\operatorname {d} \over \operatorname {d} x}\sec u=\sec u\tan u}
d d x csc u = − csc u cot u {\displaystyle {\operatorname {d} \over \operatorname {d} x}\csc u=-\csc u\cot u}
证明 sin u {\displaystyle \sin u} 的导数
令 y = sin u {\displaystyle y=\sin u} ,
则
y ′ = sin ( u + Δ u ) {\displaystyle y\prime =\sin(u+\Delta u)} ;
因此
Δ y = sin ( u + Δ u ) − sin u {\displaystyle \Delta y=\sin(u+\Delta u)-\sin u} ;
在三角学中，
sin A − sin B = 2 sin 1 2 ( A − B ) cos 1 2 ( A + B ) {\displaystyle \sin A-\sin B=2\sin {1 \over 2}(A-B)\cos {1 \over 2}(A+B)}
如果我们用 A = u + Δ u {\displaystyle A=u+\Delta u} 和 B = u {\displaystyle B=u} 代替，
我们有
Δ y = 2 cos ( u + Δ u 2 ) s i n Δ u 2 {\displaystyle \Delta y=2\cos \left(u+{\Delta u \over 2}\right)\ sin{\Delta u \over 2}}
Δ y Δ x = cos ( u + Δ u 2 ) sin Δ u 2 Δ u 2 {\displaystyle {\Delta y \over \Delta x}=\cos \left(u+{\Delta u \over 2}\right){\sin {\Delta u \over 2} \over {\Delta u \over 2}}}
当 Δ x {\displaystyle \Delta x} 趋近于零时， Δ u {\displaystyle \Delta u} 也趋近于零，并且由于 Δ u {\displaystyle \Delta u} 是用弧度表示的，则
sin Δ u 2 Δ u 2 {\displaystyle {\sin {\Delta u \over 2} \over {\Delta u \over 2}}}
d y d x = cos u {\displaystyle {\operatorname {d} y \over \operatorname {d} x}=\cos u}
证明 d d x cos u {\displaystyle {\operatorname {d} \over \operatorname {d} x}\cos u}
d d x cos u = sin u ( − d u d x ) = − sin u d u d x {\displaystyle {\operatorname {d} \over \operatorname {d} x}\cos u=\sin u\left(-{\operatorname {d} u \over \operatorname {d} x}\right)=-\sin u{\operatorname {d} u \over \operatorname {d} x}}
证明 d d x tan u {\displaystyle {\operatorname {d} \over \operatorname {d} x}\tan u}
因为 tan u = sin u cos u {\displaystyle \tan u={\sin u \over \cos u}}
d d x tan u = cos u d d x sin u − sin u d d x cos u cos 2 u {\displaystyle {\operatorname {d} \over \operatorname {d} x}\tan u={\cos u{\operatorname {d} \over \operatorname {d} x}\sin u-\sin u{\operatorname {d} \over \operatorname {d} x}\cos u \over \cos ^{2}u}}
= cos 2 u d u d x + sin 2 u d u d x cos 2 u = d u cos 2 u {\displaystyle ={\cos ^{2}u{\operatorname {d} u \over \operatorname {d} x}+\sin ^{2}u{\operatorname {d} u \over \operatorname {d} x} \over \cos ^{2}u}={\operatorname {d} u \over \cos ^{2}u}}
= sec 2 u d u d x {\displaystyle =\sec ^{2}u{\operatorname {d} u \over \operatorname {d} x}}
证明 d d x sec u {\displaystyle {\operatorname {d} \over \operatorname {d} x}\sec u}
d d x tan u = cos u d d x sin u − sin u d d x cos u cos 2 u {\displaystyle {\operatorname {d} \over \operatorname {d} x}\tan u={\cos u{\operatorname {d} \over {d}x}\sin u-\sin u{\operatorname {d} \over \operatorname {d} x}\cos u \over \cos ^{2}u}}
= cos 2 d u d x + sin 2 u d u d x cos 2 u = d u d x cos 2 u {\displaystyle ={\cos ^{2}{\operatorname {d} u \over \operatorname {d} x}+\sin ^{2}u{\operatorname {d} u \over \operatorname {d} x} \over \cos ^{2}u}={{\operatorname {d} u \over \operatorname {d} x} \over \cos ^{2}u}}
的导数
,
;
;
和 
代替，
趋近于零时，
也趋近于零，并且由于 是用弧度表示的，则
