Fungsi
Terbitan pembezaan
sin
(
x
)
{\displaystyle \sin(x)}
d
d
x
sin
(
x
)
=
kos
(
x
)
{\displaystyle {d \over dx}\sin(x)=\operatorname {kos} (x)}
kos
(
x
)
{\displaystyle \operatorname {kos} (x)}
d
d
x
kos
(
x
)
=
−
sin
(
x
)
{\displaystyle {d \over dx}\operatorname {kos} (x)=-\sin(x)}
tan
(
x
)
{\displaystyle \tan(x)}
d
d
x
tan
(
x
)
=
d
d
x
(
sin
(
x
)
kos
(
x
)
)
=
kos
2
(
x
)
+
sin
2
(
x
)
kos
2
(
x
)
=
1
+
tan
2
(
x
)
=
sek
2
(
x
)
{\displaystyle {d \over dx}\tan(x)={d \over dx}{\biggl (}{\sin(x) \over \operatorname {kos} (x)}{\biggr )}={\operatorname {kos} ^{2}(x)+\sin ^{2}(x) \over \operatorname {kos} ^{2}(x)}=1+\tan ^{2}(x)=\operatorname {sek} ^{2}(x)}
sek
(
x
)
{\displaystyle \operatorname {sek} (x)}
d
d
x
sek
(
x
)
=
d
d
x
(
1
kos
(
x
)
)
=
sin
(
x
)
kos
2
(
x
)
=
sek
(
x
)
tan
(
x
)
{\displaystyle {d \over dx}\operatorname {sek} (x)={d \over dx}{\biggl (}{1 \over \operatorname {kos} (x)}{\biggr )}={\sin(x) \over \operatorname {kos} ^{2}(x)}=\operatorname {sek} (x)\tan(x)}
kosek
(
x
)
{\displaystyle \operatorname {kosek} (x)}
d
d
x
kosek
(
x
)
=
d
d
x
(
1
sin
(
x
)
)
=
−
kos
(
x
)
sin
2
(
x
)
=
−
kotan
(
x
)
kosek
(
x
)
{\displaystyle {d \over dx}\operatorname {kosek} (x)={d \over dx}{\biggl (}{1 \over \sin(x)}{\biggr )}=-{\operatorname {kos} (x) \over \sin ^{2}(x)}=-\operatorname {kotan} (x)\operatorname {kosek} (x)}
kotan
(
x
)
{\displaystyle \operatorname {kotan} (x)}
d
d
x
kotan
(
x
)
=
d
d
x
(
kos
(
x
)
sin
(
x
)
)
=
−
kos
(
x
)
−
sin
2
sin
2
(
x
)
=
−
(
1
+
kot
2
(
x
)
)
=
−
kosek
2
(
x
)
{\displaystyle {d \over dx}\operatorname {kotan} (x)={d \over dx}{\biggl (}{\operatorname {kos} (x) \over \sin(x)}{\biggr )}={-\operatorname {kos} (x)-\sin ^{2} \over \sin ^{2}(x)}=-(1+\operatorname {kot} ^{2}(x))=-\operatorname {kosek} ^{2}(x)}
