Fungsi hiperbolik: Perbedaan antara revisi
Konten dihapus Konten ditambahkan
Baris 223:
\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 \ne 0
\end{align}</math>
<!--
Pembuktian <math>arcsinh x = ln (x + \sqrt{x^2 + 1})</math>!
: <math>y = arcsinh x</math>
: <math>x = arcsinh y</math>
: <math>x = \frac{e^y - e^{-y}}{2}</math>
: <math>2x = e^y - e^{-y}</math>
: <math>2e^yx = e^2y - 1</math>
: <math>(e^y)^2 - 2x(e^y) - 1 = 0</math>
: <math>e^y = \frac{2x + \sqrt{4x^2 + 4}}{2}</math>
: <math>e^y = x + \sqrt{x^2 + 1}</math>
: <math>ln e^y = ln (x + \sqrt{x^2 + 1})</math>
: <math>y ln e = ln (x + \sqrt{x^2 + 1})</math>
: <math>y = ln (x + \sqrt{x^2 + 1})</math>
: <math>arcsinh x = ln (x + \sqrt{x^2 + 1})</math>
-->
== Antiturunan ==
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