Soal yang Akan Dibahas
Jika $ f(x) = \frac{\cos x - \sin x}{\cos x + \sin x} $ dengan $ \cos x + \sin x \neq 0 $,
maka $ f^\prime (x) = .... $
A). $ 1 - (f(x))^2 \, $
B). $ -1 + (f(x))^2 \, $
C). $ -(1 + (f(x))^2) \, $
D). $ 1 + (f(x))^2 \, $
E). $ (f(x))^2 $
A). $ 1 - (f(x))^2 \, $
B). $ -1 + (f(x))^2 \, $
C). $ -(1 + (f(x))^2) \, $
D). $ 1 + (f(x))^2 \, $
E). $ (f(x))^2 $
$\spadesuit $ Konsep Dasar
*). Turunan Fungsi Trigonometri :
$ y = \sin x \rightarrow y^\prime = \cos x $
$ y = \cos x \rightarrow y^\prime = -\sin x $
*). Turunan Bentuk Pecahan :
$ ^\prime y = \frac{U}{V} \rightarrow y^\prime = \frac{U^\prime V - U V ^\prime }{V^2} $
*). Turunan Fungsi Trigonometri :
$ y = \sin x \rightarrow y^\prime = \cos x $
$ y = \cos x \rightarrow y^\prime = -\sin x $
*). Turunan Bentuk Pecahan :
$ ^\prime y = \frac{U}{V} \rightarrow y^\prime = \frac{U^\prime V - U V ^\prime }{V^2} $
$\clubsuit $ Pembahasan
*). Menyelesaikan soal :
$\begin{align} f(x) & = \frac{\cos x - \sin x}{\cos x + \sin x} = \frac{U}{V} \\ U & = \cos x - \sin x \rightarrow U^\prime = -\sin x - \cos x \\ V & = \cos x + \sin x \rightarrow U^\prime = -\sin x + \cos x \\ f^\prime (x) & = \frac{U^\prime V - U V ^\prime }{V^2} \\ & = \frac{(-\sin x - \cos x)(\cos x + \sin x) -(\cos x - \sin x)(-\sin x + \cos x) }{(\cos x + \sin x)^2} \\ & = \frac{-(\sin x + \cos x)(\cos x + \sin x) -(\cos x - \sin x)(\cos x -\sin x ) }{(\cos x + \sin x)^2} \\ & = \frac{-(\sin x + \cos x)^2 -(\cos x - \sin x)^2 }{(\cos x + \sin x)^2} \\ & = - \left[ \frac{(\sin x + \cos x)^2 + (\cos x - \sin x)^2 }{(\cos x + \sin x)^2} \right] \\ & = - \left[ \frac{(\sin x + \cos x)^2 }{(\cos x + \sin x)^2} + \frac{(\cos x - \sin x)^2 }{(\cos x + \sin x)^2} \right] \\ & = - \left[ 1 + \left( \frac{(\cos x - \sin x) }{(\cos x + \sin x) } \right)^2 \right] \\ & = - \left[ 1 + \left( f(x) \right)^2 \right] \end{align} $
Jadi, bentuk $ f^\prime (x) = - \left[ 1 + \left( f(x) \right)^2 \right] . \, \heartsuit $
*). Menyelesaikan soal :
$\begin{align} f(x) & = \frac{\cos x - \sin x}{\cos x + \sin x} = \frac{U}{V} \\ U & = \cos x - \sin x \rightarrow U^\prime = -\sin x - \cos x \\ V & = \cos x + \sin x \rightarrow U^\prime = -\sin x + \cos x \\ f^\prime (x) & = \frac{U^\prime V - U V ^\prime }{V^2} \\ & = \frac{(-\sin x - \cos x)(\cos x + \sin x) -(\cos x - \sin x)(-\sin x + \cos x) }{(\cos x + \sin x)^2} \\ & = \frac{-(\sin x + \cos x)(\cos x + \sin x) -(\cos x - \sin x)(\cos x -\sin x ) }{(\cos x + \sin x)^2} \\ & = \frac{-(\sin x + \cos x)^2 -(\cos x - \sin x)^2 }{(\cos x + \sin x)^2} \\ & = - \left[ \frac{(\sin x + \cos x)^2 + (\cos x - \sin x)^2 }{(\cos x + \sin x)^2} \right] \\ & = - \left[ \frac{(\sin x + \cos x)^2 }{(\cos x + \sin x)^2} + \frac{(\cos x - \sin x)^2 }{(\cos x + \sin x)^2} \right] \\ & = - \left[ 1 + \left( \frac{(\cos x - \sin x) }{(\cos x + \sin x) } \right)^2 \right] \\ & = - \left[ 1 + \left( f(x) \right)^2 \right] \end{align} $
Jadi, bentuk $ f^\prime (x) = - \left[ 1 + \left( f(x) \right)^2 \right] . \, \heartsuit $
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