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Soal Turunan Trigonometri dengan Modifikasi Identitas

Topik Bahasan ,
Soal 1. $f(x)= \frac {\sin ^2x + \cos ^2x }{\cos ^2x} \, \, \, ,f'(\frac {2 \pi}{3})=... $
Jawab:
$f(x)= \frac {\sin ^2x + \cos ^2x }{\cos ^2x} \\ f(x)= \frac {1}{\cos ^2x } = \sec^2 x \\ f'(x)= 2 \sec x. \sec x . \tan x \\ f'(\frac {2 \pi}{3})= 2 \sec (\frac {2 \pi}{3}). \sec (\frac {2 \pi}{3}) . \tan (\frac {2 \pi}{3}) \\ f'(\frac {2 \pi}{3})= 2.-2 . -2. - \sqrt 3 = - 8 \sqrt 3$

Soal 2. $f(x)= \frac {\sin ^2x + \cos ^2x }{\sin ^2x} \, \, \, ,f'(\frac {2 \pi}{3})=... $
Jawab:
$f(x)= \frac {\sin ^2x + \cos ^2x }{\sin ^2x} \\ f(x)= \frac {1}{\sin ^2 x} = \csc^2 x \\ f'(x)= 2 \csc x. (-\csc x . \cot x) \\ f'(x)= -2 \csc x. \csc x . \cot x \\ f'(\frac {2 \pi}{3})=-2 \csc (\frac {2 \pi}{3}). \csc (\frac {2 \pi}{3}) . \cot (\frac {2 \pi}{3}) \\  -2.\frac {2}{\sqrt 3} . \frac {2}{\sqrt 3}. - \frac {1}{\sqrt 3} =\frac {8}{9} \sqrt 3$.

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