Fungsi hiperbolik

Hyperbolic functions-2

Fungsi hiperbolik adalah salah satu hasil kombinasi dari fungsi-fungsi eksponen. Fungsi hiperbolik memiliki rumus. Selain itu memiliki invers serta turunan dan anti turunan fungsi hiperbolik dan inversnya.

Definisi

Definisi Eksponen

Dalam istilah dari fungsi eksponensial:

• Hiperbolik sinus:

  $\sinh x=\frac{e^x-e^{-x}}{2}=\frac{e^{2x}-1}{2e^x}=\frac{1-e^{-2x}}{2e^{-x}}.$

• Hiperbolik kosinus:

  $\cosh x=\frac{e^x+e^{-x}}{2}=\frac{e^{2x}+1}{2e^x}=\frac{1+e^{-2x}}{2e^{-x}}.$

• Hiperbolik tangen:

  $\tanh x=\frac{\sinh x}{\cosh x}=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^{2x}-1}{e^{2x}+1}$

• Hiperbolik kotangen: untuk ,

  $\coth x=\frac{\cosh x}{\sinh x}=\frac{e^x+e^{-x}}{e^x-e^{-x}}=\frac{e^{2x}+1}{e^{2x}-1}$

• Hiperbolik sekan:

  $\mathrm{sech}x=\frac{1}{\cosh x}=\frac{2}{e^x+e^{-x}}=\frac{2e^x}{e^{2x}+1}$

• Hiperbolik kosekan: untuk ,

  $\mathrm{csch}x=\frac{1}{\sinh x}=\frac{2}{e^x-e^{-x}}=\frac{2e^x}{e^{2x}-1}$

Definisi persamaan diferensial

- Dalam pengembangan -

Definisi kompleks trigonometri

-Dalam pengembangan -

Sifat karakteristik

- Dalam pengembangan -

Penambahan

$\begin{array}{cc}\sinh (x+y)& =\sinh x\cosh y+\cosh x\sinh y\\ \cosh (x+y)& =\cosh x\cosh y+\sinh x\sinh y\\ [6px]\tanh (x+y)& =\frac{\tanh x+\tanh y}{1+\tanh x\tanh y}\\ \end{array}$

terutama

$\begin{array}{cc}\cosh (2x)& ={\sinh }^{2}x+{\cosh }^{2}x=2{\sinh }^{2}x+1=2{\cosh }^{2}x-1\\ \sinh (2x)& =2\sinh x\cosh x\\ \tanh (2x)& =\frac{2\tanh x}{1+{\tanh }^{2}x}\\ \end{array}$

Lihat:

$\begin{array}{cc}\sinh x+\sinh y& =2\sinh \left(\frac{x+y}{2}\right)\cosh \left(\frac{x-y}{2}\right)\\ \cosh x+\cosh y& =2\cosh \left(\frac{x+y}{2}\right)\cosh \left(\frac{x-y}{2}\right)\\ \end{array}$

Pengurangan

$\begin{array}{cc}\sinh (x-y)& =\sinh x\cosh y-\cosh x\sinh y\\ \cosh (x-y)& =\cosh x\cosh y-\sinh x\sinh y\\ \tanh (x-y)& =\frac{\tanh x-\tanh y}{1-\tanh x\tanh y}\\ \end{array}$

Dan juga:

$\begin{array}{cc}\sinh x-\sinh y& =2\cosh \left(\frac{x+y}{2}\right)\sinh \left(\frac{x-y}{2}\right)\\ \cosh x-\cosh y& =2\sinh \left(\frac{x+y}{2}\right)\sinh \left(\frac{x-y}{2}\right)\\ \end{array}$

Rumus setengah argumen

$\begin{array}{cccc}\sinh \left(\frac{x}{2}\right)& =\frac{\sinh x}{\sqrt{2(\cosh x+1)}}& & =\mathrm{sgn}x\sqrt{\frac{\cosh x-1}{2}}\\ [6px]\cosh \left(\frac{x}{2}\right)& =\sqrt{\frac{\cosh x+1}{2}}\\ [6px]\tanh \left(\frac{x}{2}\right)& =\frac{\sinh x}{\cosh x+1}& & =\mathrm{sgn}x\sqrt{\frac{\cosh x-1}{\cosh x+1}}=\frac{e^x-1}{e^x+1}\end{array}$

di mana adalah fungsi tanda.

Jika $x\ne 0$, maka

$\tanh \left(\frac{x}{2}\right)=\frac{\cosh x-1}{\sinh x}=\coth x-\mathrm{csch}x$

Rumus kuadrat

$\begin{array}{cc}{\sinh }^{2}x& =\frac{1}{2}(\cosh 2x-1)\\ {\cosh }^{2}x& =\frac{1}{2}(\cosh 2x+1)\end{array}$

Pertidaksamaan

Pertidaksamaan berikut sangat berguna dalam statistik, yaitu $\cosh (t)\le e^{t^2/2}$

Fungsi invers sebagai logaritma

$\begin{array}{cc}\mathrm{arsinh}(x)& =\ln (x+\sqrt{x^2+1})\\ \mathrm{arcosh}(x)& =\ln (x+\sqrt{x^2-1})& & x⩾1\\ \mathrm{artanh}(x)& =\frac{1}{2}\ln \left(\frac{1+x}{1-x}\right)& & |x|<1\\ \mathrm{arcoth}(x)& =\frac{1}{2}\ln \left(\frac{x+1}{x-1}\right)& & |x|>1\\ \mathrm{arsech}(x)& =\ln (\frac{1}{x}+\sqrt{\frac{1}{x^2}-1})=\ln \left(\frac{1+\sqrt{1-x^2}}{x}\right)& & 0<x⩽1\\ \mathrm{arcsch}(x)& =\ln (\frac{1}{x}+\sqrt{\frac{1}{x^2}+1})=\ln \left(\frac{1+\sqrt{1+x^2}}{x}\right)& & x\ne 0\end{array}$

Turunan

$\begin{array}{cc}\frac{d}{dx}\sinh x& =\cosh x\\ \frac{d}{dx}\cosh x& =\sinh x\\ \frac{d}{dx}\tanh x& =1-{\tanh }^{2}x={\mathrm{sech}}^{2}x=\frac{1}{{\cosh }^{2}x}\\ \frac{d}{dx}\coth x& =1-{\coth }^{2}x=-{\mathrm{csch}}^{2}x=-\frac{1}{{\sinh }^{2}x}& & x\ne 0\\ \frac{d}{dx}\mathrm{sech}x& =-\tanh x\mathrm{sech}x\\ \frac{d}{dx}\mathrm{csch}x& =-\coth x\mathrm{csch}x& & x\ne 0\\ \frac{d}{dx}\mathrm{arsinh}x& =\frac{1}{\sqrt{x^2+1}}\\ \frac{d}{dx}\mathrm{arcosh}x& =\frac{1}{\sqrt{x^2-1}}& & 1<x\\ \frac{d}{dx}\mathrm{artanh}x& =\frac{1}{1-x^2}& & |x|<1\\ \frac{d}{dx}\mathrm{arcoth}x& =\frac{1}{1-x^2}& & 1<|x|\\ \frac{d}{dx}\mathrm{arsech}x& =-\frac{1}{x\sqrt{1-x^2}}& & 0<x<1\\ \frac{d}{dx}\mathrm{arcsch}x& =-\frac{1}{|x|\sqrt{1+x^2}}& & x\ne 0\end{array}$

Turunan detik

- Dalam pengembangan -

Standar integral

$\begin{array}{cc}\int \sinh (ax)dx& =a^{-1}\cosh (ax)+C\\ \int \cosh (ax)dx& =a^{-1}\sinh (ax)+C\\ \int \tanh (ax)dx& =a^{-1}\ln (\cosh (ax))+C\\ \int \coth (ax)dx& =a^{-1}\ln (\sinh (ax))+C\\ \int \mathrm{sech}(ax)dx& =a^{-1}\arctan (\sinh (ax))+C\\ \int \mathrm{csch}(ax)dx& =a^{-1}\ln (\tanh \left(\frac{ax}{2}\right))+C=a^{-1}\ln |\mathrm{csch}(ax)-\coth (ax)|+C\end{array}$

---

$\begin{array}{cc}\int \frac{1}{\sqrt{a^2+u^2}}du& =\mathrm{arsinh}\left(\frac{u}{a}\right)+C\\ \int \frac{1}{\sqrt{u^2-a^2}}du& =\mathrm{arcosh}\left(\frac{u}{a}\right)+C\\ \int \frac{1}{a^2-u^2}du& =a^{-1}\mathrm{artanh}\left(\frac{u}{a}\right)+C& & u^2<a^2\\ \int \frac{1}{a^2-u^2}du& =a^{-1}\mathrm{arcoth}\left(\frac{u}{a}\right)+C& & u^2>a^2\\ \int \frac{1}{u\sqrt{a^2-u^2}}du& =-a^{-1}\mathrm{arsech}\left(\frac{u}{a}\right)+C\\ \int \frac{1}{u\sqrt{a^2+u^2}}du& =-a^{-1}\mathrm{arcsch}\left|\frac{u}{a}\right|+C\end{array}$

Sumber dan atribusi

Konten artikel ini diadaptasi dari Wikipedia bahasa Indonesia, revisi 23907604 (2023-07-26T13:41:25Z), yang tersedia berdasarkan lisensi Creative Commons Atribusi-BerbagiSerupa (CC BY-SA). Gambar pada artikel ini bersumber dari Wikimedia Commons dan mengikuti ketentuan lisensi masing-masing berkas. Mohon gunakan konten dan media secara bijak serta sesuai dengan ketentuan lisensi yang berlaku.