Pembahasan Turunan SBMPTN 2017 Matematika IPA kode 168

Soal yang Akan Dibahas
Misalkan $ f(x) = 2\tan \left( \sqrt{\sec x} \right) $ , maka $ f^\prime (x) = .... $
A). $ \sec ^2 \left( \sqrt{\sec x} \right) . \tan x \, $
B). $ \sec ^2 \left( \sqrt{\sec x} \right) . \sqrt{\sec x}. \tan x \, $
C). $ 2\sec ^2 \left( \sqrt{\sec x} \right) . \sqrt{\sec x} . \tan x \, $
D). $ \sec ^2 \left( \sqrt{\sec x} \right) . \sec x . \tan x \, $
E). $ 2\sec ^2 \left( \sqrt{\sec x} \right) . \sec x . \tan x $

$\spadesuit $ Konsep Dasar
*). Turunan fungsi trigonometri :
$ y = \tan g(x) \rightarrow y^\prime = \sec ^2 ( g(x) ) . g^\prime (x) $.
$ y = \sec x \rightarrow y^\prime = \sec x . \tan x $
$ y = [h(x)]^n \rightarrow y^\prime = n[h(x)]^{n-1}. h^\prime (x) $.
*). Sifat eksponen : $ a^m.a^n = a^{m+n} $ dan $ \sqrt{a} = a^\frac{1}{2} $

$\clubsuit $ Pembahasan
*). Misalkan $ g(x) = \sqrt{\sec x } = (\sec x )^\frac{1}{2} $ , Turunannya :
$ \begin{align} g^\prime (x) & = \frac{1}{2} (\sec x)^{-\frac{1}{2}} . \sec x . \tan x \\ & = \frac{1}{2} (\sec x)^{-\frac{1}{2} + 1} . \tan x \\ & = \frac{1}{2} (\sec x)^{\frac{1}{2} } . \tan x \\ & = \frac{1}{2} \sqrt{ \sec x} . \tan x \end{align} $
*). Menyelesaikan soal :
$\begin{align} f(x) & = 2\tan \left( \sqrt{\sec x} \right) \\ f(x) & = 2\tan \left( g(x) \right) \\ f^\prime (x) & = 2\sec ^2 ( g(x) ) . g^\prime (x) \\ f^\prime (x) & = 2\sec ^2 \left( \sqrt{\sec x} \right) . \frac{1}{2} \sqrt{ \sec x} . \tan x \\ f^\prime (x) & = \sec ^2 \left( \sqrt{\sec x} \right) . \sqrt{ \sec x} . \tan x \end{align} $
Jadi, $ f^\prime (x) = \sec ^2 \left( \sqrt{\sec x} \right) . \sqrt{ \sec x} . \tan x . \, \heartsuit $

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