Fungsi Hiperbolik adalah salah satu hasil kombinasi dari fungsi-fungsi eksponen .[ 1] Fungsi Hiperbolik memiliki rumus atau formula.[ 1] Selain itu memiliki invers serta turunan dan anti turunan fungsi hiperbolik dan inversnya. e x e− x .[ 1]
Fungsi hiperbolik
Definisi
Properti Karakteristik
- Dalam pengembangan -
Argumen produk
sinh
(
x
+
y
)
=
sinh
x
cosh
y
+
cosh
x
sinh
y
cosh
(
x
+
y
)
=
cosh
x
cosh
y
+
sinh
x
sinh
y
tanh
(
x
+
y
)
=
tanh
x
+
tanh
y
1
+
tanh
x
tanh
y
{\displaystyle {\begin{aligned}\sinh(x+y)&=\sinh x\cosh y+\cosh x\sinh y\\\cosh(x+y)&=\cosh x\cosh y+\sinh x\sinh y\\[6px]\tanh(x+y)&={\frac {\tanh x+\tanh y}{1+\tanh x\tanh y}}\\\end{aligned}}}
terutama
cosh
(
2
x
)
=
sinh
2
x
+
cosh
2
x
=
2
sinh
2
x
+
1
=
2
cosh
2
x
−
1
sinh
(
2
x
)
=
2
sinh
x
cosh
x
tanh
(
2
x
)
=
2
tanh
x
1
+
tanh
2
x
{\displaystyle {\begin{aligned}\cosh(2x)&=\sinh ^{2}{x}+\cosh ^{2}{x}=2\sinh ^{2}x+1=2\cosh ^{2}x-1\\\sinh(2x)&=2\sinh x\cosh x\\\tanh(2x)&={\frac {2\tanh x}{1+\tanh ^{2}x}}\\\end{aligned}}}
Lihat:
sinh
x
+
sinh
y
=
2
sinh
(
x
+
y
2
)
cosh
(
x
−
y
2
)
cosh
x
+
cosh
y
=
2
cosh
(
x
+
y
2
)
cosh
(
x
−
y
2
)
{\displaystyle {\begin{aligned}\sinh x+\sinh y&=2\sinh \left({\frac {x+y}{2}}\right)\cosh \left({\frac {x-y}{2}}\right)\\\cosh x+\cosh y&=2\cosh \left({\frac {x+y}{2}}\right)\cosh \left({\frac {x-y}{2}}\right)\\\end{aligned}}}
Rumus
sinh
(
x
−
y
)
=
sinh
x
cosh
y
−
cosh
x
sinh
y
cosh
(
x
−
y
)
=
cosh
x
cosh
y
−
sinh
x
sinh
y
tanh
(
x
−
y
)
=
tanh
x
−
tanh
y
1
−
tanh
x
tanh
y
{\displaystyle {\begin{aligned}\sinh(x-y)&=\sinh x\cosh y-\cosh x\sinh y\\\cosh(x-y)&=\cosh x\cosh y-\sinh x\sinh y\\\tanh(x-y)&={\frac {\tanh x-\tanh y}{1-\tanh x\tanh y}}\\\end{aligned}}}
Lihat:[ 2]
sinh
x
−
sinh
y
=
2
cosh
(
x
+
y
2
)
sinh
(
x
−
y
2
)
cosh
x
−
cosh
y
=
2
sinh
(
x
+
y
2
)
sinh
(
x
−
y
2
)
{\displaystyle {\begin{aligned}\sinh x-\sinh y&=2\cosh \left({\frac {x+y}{2}}\right)\sinh \left({\frac {x-y}{2}}\right)\\\cosh x-\cosh y&=2\sinh \left({\frac {x+y}{2}}\right)\sinh \left({\frac {x-y}{2}}\right)\\\end{aligned}}}
Rumus argumen tinggi
sinh
(
x
2
)
=
sinh
x
2
(
cosh
x
+
1
)
=
sgn
x
cosh
x
−
1
2
cosh
(
x
2
)
=
cosh
x
+
1
2
tanh
(
x
2
)
=
sinh
x
cosh
x
+
1
=
sgn
x
cosh
x
−
1
cosh
x
+
1
=
e
x
−
1
e
x
+
1
{\displaystyle {\begin{aligned}\sinh \left({\frac {x}{2}}\right)&={\frac {\sinh x}{\sqrt {2(\cosh x+1)}}}&&=\operatorname {sgn} x\,{\sqrt {\frac {\cosh x-1}{2}}}\\[6px]\cosh \left({\frac {x}{2}}\right)&={\sqrt {\frac {\cosh x+1}{2}}}\\[6px]\tanh \left({\frac {x}{2}}\right)&={\frac {\sinh x}{\cosh x+1}}&&=\operatorname {sgn} x\,{\sqrt {\frac {\cosh x-1}{\cosh x+1}}}={\frac {e^{x}-1}{e^{x}+1}}\end{aligned}}}
Darimana sgn adalah tanda fungsi .
If x ≠ 0 , then[ 3]
tanh
(
x
2
)
=
cosh
x
−
1
sinh
x
=
coth
x
−
csch
x
{\displaystyle \tanh \left({\frac {x}{2}}\right)={\frac {\cosh x-1}{\sinh x}}=\coth x-\operatorname {csch} x}
Rumus persegi
sinh
2
x
=
1
2
(
cosh
2
x
−
1
)
cosh
2
x
=
1
2
(
cosh
2
x
+
1
)
{\displaystyle {\begin{aligned}\sinh ^{2}x&={\frac {1}{2}}(\cosh 2x-1)\\\cosh ^{2}x&={\frac {1}{2}}(\cosh 2x+1)\end{aligned}}}
Pertidaksamaan
Jika Pertidaksamaan saat ada statistik yaitu:
cosh
(
t
)
≤
e
t
2
/
2
{\displaystyle \operatorname {cosh} (t)\leq e^{t^{2}/2}}
[ 4]
Fungsi invers sebagai logaritma
arsinh
(
x
)
=
ln
(
x
+
x
2
+
1
)
arcosh
(
x
)
=
ln
(
x
+
x
2
−
1
)
x
⩾
1
artanh
(
x
)
=
1
2
ln
(
1
+
x
1
−
x
)
|
x
|
<
1
arcoth
(
x
)
=
1
2
ln
(
x
+
1
x
−
1
)
|
x
|
>
1
arsech
(
x
)
=
ln
(
1
x
+
1
x
2
−
1
)
=
ln
(
1
+
1
−
x
2
x
)
0
<
x
⩽
1
arcsch
(
x
)
=
ln
(
1
x
+
1
x
2
+
1
)
x
≠
0
{\displaystyle {\begin{aligned}\operatorname {arsinh} (x)&=\ln \left(x+{\sqrt {x^{2}+1}}\right)\\\operatorname {arcosh} (x)&=\ln \left(x+{\sqrt {x^{2}-1}}\right)&&x\geqslant 1\\\operatorname {artanh} (x)&={\frac {1}{2}}\ln \left({\frac {1+x}{1-x}}\right)&&|x|<1\\\operatorname {arcoth} (x)&={\frac {1}{2}}\ln \left({\frac {x+1}{x-1}}\right)&&|x|>1\\\operatorname {arsech} (x)&=\ln \left({\frac {1}{x}}+{\sqrt {{\frac {1}{x^{2}}}-1}}\right)=\ln \left({\frac {1+{\sqrt {1-x^{2}}}}{x}}\right)&&0<x\leqslant 1\\\operatorname {arcsch} (x)&=\ln \left({\frac {1}{x}}+{\sqrt {{\frac {1}{x^{2}}}+1}}\right)&&x\neq 0\end{aligned}}}
Antiturunan
d
d
x
sinh
x
=
cosh
x
d
d
x
cosh
x
=
sinh
x
d
d
x
tanh
x
=
1
−
tanh
2
x
=
sech
2
x
=
1
cosh
2
x
d
d
x
coth
x
=
1
−
coth
2
x
=
−
csch
2
x
=
−
1
sinh
2
x
x
≠
0
d
d
x
sech
x
=
−
tanh
x
sech
x
d
d
x
csch
x
=
−
coth
x
csch
x
x
≠
0
d
d
x
arsinh
x
=
1
x
2
+
1
d
d
x
arcosh
x
=
1
x
2
−
1
1
<
x
d
d
x
artanh
x
=
1
1
−
x
2
|
x
|
<
1
d
d
x
arcoth
x
=
1
1
−
x
2
1
<
|
x
|
d
d
x
arsech
x
=
−
1
x
1
−
x
2
0
<
x
<
1
d
d
x
arcsch
x
=
−
1
|
x
|
1
+
x
2
x
≠
0
{\displaystyle {\begin{aligned}{\frac {d}{dx}}\sinh x&=\cosh x\\{\frac {d}{dx}}\cosh x&=\sinh x\\{\frac {d}{dx}}\tanh x&=1-\tanh ^{2}x=\operatorname {sech} ^{2}x={\frac {1}{\cosh ^{2}x}}\\{\frac {d}{dx}}\coth x&=1-\coth ^{2}x=-\operatorname {csch} ^{2}x=-{\frac {1}{\sinh ^{2}x}}&&x\neq 0\\{\frac {d}{dx}}\operatorname {sech} x&=-\tanh x\operatorname {sech} x\\{\frac {d}{dx}}\operatorname {csch} x&=-\coth x\operatorname {csch} x&&x\neq 0\\{\frac {d}{dx}}\operatorname {arsinh} x&={\frac {1}{\sqrt {x^{2}+1}}}\\{\frac {d}{dx}}\operatorname {arcosh} x&={\frac {1}{\sqrt {x^{2}-1}}}&&1<x\\{\frac {d}{dx}}\operatorname {artanh} x&={\frac {1}{1-x^{2}}}&&|x|<1\\{\frac {d}{dx}}\operatorname {arcoth} x&={\frac {1}{1-x^{2}}}&&1<|x|\\{\frac {d}{dx}}\operatorname {arsech} x&=-{\frac {1}{x{\sqrt {1-x^{2}}}}}&&0<x<1\\{\frac {d}{dx}}\operatorname {arcsch} x&=-{\frac {1}{|x|{\sqrt {1+x^{2}}}}}&&x\neq 0\end{aligned}}}
Antiturunan detik
Standar integral
∫
sinh
(
a
x
)
d
x
=
a
−
1
cosh
(
a
x
)
+
C
∫
cosh
(
a
x
)
d
x
=
a
−
1
sinh
(
a
x
)
+
C
∫
tanh
(
a
x
)
d
x
=
a
−
1
ln
(
cosh
(
a
x
)
)
+
C
∫
coth
(
a
x
)
d
x
=
a
−
1
ln
(
sinh
(
a
x
)
)
+
C
∫
sech
(
a
x
)
d
x
=
a
−
1
arctan
(
sinh
(
a
x
)
)
+
C
∫
csch
(
a
x
)
d
x
=
a
−
1
ln
(
tanh
(
a
x
2
)
)
+
C
=
a
−
1
ln
|
csch
(
a
x
)
−
coth
(
a
x
)
|
+
C
{\displaystyle {\begin{aligned}\int \sinh(ax)\,dx&=a^{-1}\cosh(ax)+C\\\int \cosh(ax)\,dx&=a^{-1}\sinh(ax)+C\\\int \tanh(ax)\,dx&=a^{-1}\ln(\cosh(ax))+C\\\int \coth(ax)\,dx&=a^{-1}\ln(\sinh(ax))+C\\\int \operatorname {sech} (ax)\,dx&=a^{-1}\arctan(\sinh(ax))+C\\\int \operatorname {csch} (ax)\,dx&=a^{-1}\ln \left(\tanh \left({\frac {ax}{2}}\right)\right)+C=a^{-1}\ln \left|\operatorname {csch} (ax)-\coth(ax)\right|+C\end{aligned}}}
∫
1
a
2
+
u
2
d
u
=
arsinh
(
u
a
)
+
C
∫
1
u
2
−
a
2
d
u
=
arcosh
(
u
a
)
+
C
∫
1
a
2
−
u
2
d
u
=
a
−
1
artanh
(
u
a
)
+
C
u
2
<
a
2
∫
1
a
2
−
u
2
d
u
=
a
−
1
arcoth
(
u
a
)
+
C
u
2
>
a
2
∫
1
u
a
2
−
u
2
d
u
=
−
a
−
1
arsech
(
u
a
)
+
C
∫
1
u
a
2
+
u
2
d
u
=
−
a
−
1
arcsch
|
u
a
|
+
C
{\displaystyle {\begin{aligned}\int {{\frac {1}{\sqrt {a^{2}+u^{2}}}}\,du}&=\operatorname {arsinh} \left({\frac {u}{a}}\right)+C\\\int {{\frac {1}{\sqrt {u^{2}-a^{2}}}}\,du}&=\operatorname {arcosh} \left({\frac {u}{a}}\right)+C\\\int {\frac {1}{a^{2}-u^{2}}}\,du&=a^{-1}\operatorname {artanh} \left({\frac {u}{a}}\right)+C&&u^{2}<a^{2}\\\int {\frac {1}{a^{2}-u^{2}}}\,du&=a^{-1}\operatorname {arcoth} \left({\frac {u}{a}}\right)+C&&u^{2}>a^{2}\\\int {{\frac {1}{u{\sqrt {a^{2}-u^{2}}}}}\,du}&=-a^{-1}\operatorname {arsech} \left({\frac {u}{a}}\right)+C\\\int {{\frac {1}{u{\sqrt {a^{2}+u^{2}}}}}\,du}&=-a^{-1}\operatorname {arcsch} \left|{\frac {u}{a}}\right|+C\end{aligned}}}
Rumus
Fungsi hiperbolik dibangun oleh dua fungsi p dan q dengan p:
R → R+, 2 ( ) p x = ex dan q:R → R+, 2 ( ) q x e x − = .[ 1]
Selanjutnya dibangun fungsi f dan g yang dinyatakan sebagai jumlah dan selisih dari fungsi p dan q, dengan demikian:
f (x) = p(x) + q(x) dan g(x) = p(x) − q(x).[ 1] Sifat-sifat yang dimiliki oleh fungsi f dan g memiliki kemiripan dengan sifat-sifat fungsi trigonometri , salah satunya adalah kesamaan dasar fungsi yang memiliki kemiripan dengan sifat pada fungsi trigonometri .[ 1]
Dengan mengacu pada sifat-sifat tersebut, kemudian dikembangkan suatu ide untuk menyatakan fungsi f dan g sebagai fungsi hiperbolik. f 2 (x) − g 2 (x) = 1 cos 2 x + sin 2 x = 1.[ 1]
Kemudian fungsi sinus hiperbolik dan tangen hiperbolik mempunyai invers karena kedua fungsi tersebut satu-satu pada setiap daerah asalnya.[ 5] Fungsi cosinus hiperbolik tidak mempunyai invers karena fungsi ini tidak satu-satu, akan tetapi dengan membatasi daerah asal x lebih dari sama dengan 0 fungsi cosinus hiperbolik mempunyai invers.[ 5]
Referensi