C. Turunan Fungsi Invers Tan (x)
1) Invers Fungsi Tan (x)
$y\:=\:tan\: x\: \Leftrightarrow\: x\:=\:arctan\: y,$
invers $y\:=\:arctan\: x.$
2) Turunan Invers Fungsi Tan (x)
$y\:=\:arctan\: x\: \Leftrightarrow\: y’\:=\: \frac{dy}{dx}\:=\:\frac{1}{1\:+\:x^{2}}$
Pembuktian:
$y\:=\:arctan\: x\: \Leftrightarrow\: x\:=\: tan\: y$
$x\:=\: tan\: y$
kedua ruas diturunkan
$dx\:=\: sec^{2}\:y\: dy$
$\frac{dy}{dx}\: =\: \frac{1}{ sec^{2}\:y }\: =\: \frac{1}{1\:+\: tan^{2}\:y}$
$tan\: y\: =\: x$, maka
$\frac{dy}{dx}\: =\: \frac{1}{1\:+ x^{2}}$
$\therefore\: y\:=\:arctan\:(x)$
$y’\:=\: \frac{dy}{dx}\: =\: \frac{1}{1\:+\:x^{2}}$ (terbukti)
$y\:=\:arctan\:(u(x))$
$y’\:=\: \frac{dy}{dx}\: =\: \frac{u’(x)}{\sqrt{1\:+\:u(x)^{2}}}$
D. Turunan Fungsi Invers Cot (x)
1) Invers Fungsi Cot (x)
$y\:=\:cot\: x\: \Leftrightarrow \:x\:=\:arccot\: y$,
invers $y\:=\:arccot\:x$
2) Turunan Invers Fungsi Cot (x)
$y\: =\:arccot\: x\: \Leftrightarrow\: y’\:=\: \frac{dy}{dx}\:=\:- \frac{1}{1\:+\:x^{2}}$
Pembuktian:
$y\: =\: arccot\: x\: \Leftrightarrow\: x\:=\: cot\: y,\: cot\: y\:=\: \frac{1}{tan\: y}$
$x\: =\: cot\: y\: =\: \frac{1}{tan\: y},\: tan\: y\: =\: \frac{1}{x}$
sehingga
$y \:=\:arccot\: x\: =\:arctan\: \frac{1}{x}$
menggunakan sifat tangen diperoleh
$y’\: =\: \frac{-\frac{1}{x^{2}}}{1\:+\:\frac{1}{x^{2}}}$
$y’\: =\: -\frac{1}{x^{2}}. \frac{x^{2}}{ \frac{1}{x^{2}}\:+\:1}$
$y’\: =\: -\frac{1}{1\:+\:x^{2}}$
$\therefore \:y\:=\:arccot\: (x)$
$y’\:=\: \frac{dy}{dx}\:=\:- \frac{1}{1\:+\:x^{2}}$ (terbukti)
$\therefore \:y\:=\:arccot\:(u(x))$
$y’\:=\: \frac{dy}{dx}\: = \:-\frac{u’(x)}{\sqrt{1\:+\:u(x)^{2}}}$
E. Turunan Fungsi Invers Sec (x)
1) Invers Fungsi Sec (x)
$y\:=\:sec\: x\: \Leftrightarrow \:x\:=\:arcsec\: y$,
invers $y\:=\:arcsec\:x$
2) Turunan Invers Fungsi Sec (x)
$y\: =\: arcsec x\: \Leftrightarrow\: y’\: =\: \frac{1}{\mid x \mid \sqrt{x^{2}\:-\:1}}$
Pembuktian:
$y\: =\: arcsec\: x \Leftrightarrow x\: =\: sec\: y, sec\: y\: =\: \frac{1}{cos\: y}$
$y\: =\: arccos\: \frac{1}{x} $
menggunakan turunan cos u, maka diperoleh:
$y’\: =\: \frac{- \frac{1}{x^{2}}}{- \sqrt{1\: -\: \frac{1}{x}^{2}}}$
$y’\: = \:\frac{ \mid x \mid }{x^{2} . \sqrt{x^{2}\:-\:1}}$
$y’\: = \:\frac{1}{\mid x \mid \sqrt{x^{2}\:-\:1}}$, dengan syarat $\mid x \mid > 1$
$\therefore y \:=\: arcsec\: x$
$y’\: =\: \frac{1}{\mid x \mid \sqrt{x^{2}\:-\:1}}$(terbukti)
$\therefore\: y\: =\: arcsec\: u(x)$
$y’\: =\: \frac{u’}{\mid u \mid \sqrt{u^{2}\:-\:1}}$(terbukti)
F. Turunan Fungsi Invers Cosec (x)
1) Invers Fungsi Cosec (x)
$y\:=\:cosec\: x\: \Leftrightarrow \:x\:=\:arccosec\: y$,
invers $y\:=\:arccosec\:x$
2) Turunan Invers Fungsi Cosec (x)
$y\: =\: arccosec\: x\: \Leftrightarrow \:y’\: = \: -\frac{1}{\mid x \mid \sqrt{x^{2}\:-\:1}}$
Pembuktian:
$y \:=\: arccosec\: x\: \Leftrightarrow\: x\: =\: cosec\: y,\: cosec\: y\: =\: \frac{1}{sin\: y}$
$y \:=\: arcsin\: \frac{1}{x}$
menggunakan turunan sin u, maka diperoleh:
$y’\: =\: - \frac{1}{x^{2}} . \frac{1}{\sqrt{1\:-\: \frac{1}{x^{2}}}}$
$y’\: = \: -\frac{1}{\mid x \mid \sqrt{x^{2}\:-\:1}}$
$\therefore y\: = \:arcsec\: x$
$y’\: =\: -\frac{1}{\mid x \mid \sqrt{x^{2}\:-\:1}}$(terbukti)
$\therefore y \:=\: arcsec \:u(x)$
$y’\: = \:-\frac{u’}{\mid u \mid \sqrt{u^{2}\:-\:1}} $
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