Soal Turunan Trigonometri, Bentuk Pangkat dan Akar

Soal 1.  f(x)=  $3 \sqrt[3]{\sin ^2 (x+10)}$  maka $f'(50^o)$=...
Jawab:
 $f(x)=3.\sqrt[3]{\sin ^2 (x+10)} \\ f(x)= (\sin (x+10))^{\frac {2}{3}} \\ f'(x)= 3.   \frac {2}{3} (\sin (x+10))^{-\frac {1}{3}}. \cos (x+10) \\ f'(50^\circ)=2 (\sin 60^\circ ) ^{-\frac {1}{3}}. \cos 60^ \circ = \frac {1}{ \sqrt [3] {sin 60^ \circ}}$

Soal 2. $f(x)= \frac {\sin 2x + \cos 2x }{\cos 2x} \, \, \, ,f'(\frac {\pi}{6})=...$
Jawab:
$f(x)= \frac {\sin 2x + \cos 2x }{\cos 2x} \\ f(x)= \frac {\sin 2x }{\cos 2x} +1 \\ f(x)= \tan 2x+1 \\ f'(x)= 2.\sec^2 2x \\ f'(\frac {\pi}{6})=2. \sec^2 \frac {\pi}{6} \\f'(\frac {\pi}{6})= 2. (\frac {2}{\sqrt 3})^2 = \frac {8}{3}$