Soal dan Pembahasan Turunan Trigonometri (perkalian dan pecahan)

Soal 1. f(x) = 3x$^2$ sin (2x+1)... f'(x)=...
Jawab:
Bentuk perkalian gunakan rumus turunan untuk perkalian dimana,

f(x)=u.v maka f'(x)= u'v+uv'

u=3x$^2$  => u' = 6x
v= sin (2x+1)   => v' = 2 cos (2x+1)
f'(x)=   6x sin (2x+1) + 3x$^2$ 2 cos (2x+1)= 3x (2sin (2x+1)+x cos (2x+1)

Soal 2,$f(x)= \frac {6x}{\cos (6x-1)} \, \text {maka } \, f'(x)=...$
Jawab:
Fungsi berbentuk pecahan sehingga digunakan rumus turunan untuk fungsi pecahan.

$f(x)= \frac {u}{v} \Leftrightarrow  f'(x)= \frac {u'v-uv'}{v^2}$

u= 6x    u'=6
v= cos (6x-1)   v' = -6sin (6x-1)
Sesuai rumus bisa ditulis:
 $f'(x)= \frac {6.\cos (6x-1)-6x (-6 \sin (6x-1))}{(\cos (6x-1))^2} \\ f'(x)=\frac {6.\cos (6x-1)+6x .6 \sin (6x-1)}{(\cos (6x-1))^2} $

Soal 3. $f(x)= \frac {2}{\sin x \cos x} , \,\, \text  {maka } f'(x)=...$
Jawab:
$f(x)= \frac {2}{\sin x \cos x} . \frac {2}{2} = \frac {4}{2 \sin x \cos x} \\ f(x) = \frac {4}{\sin 2x} = 4 \csc 2x \\ f'(x)= 4.2 \csc 2x \cot 2x = 8 \csc 2x \cot 2x$

Soal 4. $f(x)= \frac {2}{\sin x \cos x} , \,\, \text  {maka } f'(x)=...$
Jawab:
$f(x)= \frac {2}{\sin x \cos x} . \frac {2}{2} = \frac {4}{2 \sin x \cos x} \\ f(x) = \frac {4}{\sin 2x} = 4 \csc 2x \\ f'(x)= 4.2 \csc 2x \cot 2x = 8 \csc 2x \cot 2x$

Soal 5$f(x)= \frac {\cos 2x}{1- \sin 2x} $ Maka $f (\frac {\pi}{4})=...$
Jawab:
Soal ini berbentuk pecahan, gunakan rumusan seperti nomer 2.
u= cos 2x   u'= -2sin 2x
v= 1-sin 2x    v' =-2cos 2x
Disusun sesuai rumus turunan untuk pecahan:
$f'(x)= \frac {-2\sin 2x.(1-\sin 2x)-\cos 2x (-2\cos 2x)}{(1-\sin 2x )^2} \\ f'(x)=\frac {2 \sin^2 2x-2 \sin 2x+2 \cos ^2 2x}{(1-\sin 2x )^2} \\ f'(x)=\frac {2-2 \sin 2x}{(1-\sin 2x )^2} \\ f (\frac {\pi}{4})= \frac {2-2}{(1-1)^2} = \frac {0}{0}$

Soal 6. $f(x)= \frac { \cos 2x}{\sin x + \cos x}$  Nilai dari $f'(\frac {\pi}{2})=...$
 Jawab:
f(x)= \frac { \cos 2x}{\sin x + \cos x} \\ f(x) =\frac { \cos ^2 x - \sin ^2 x}{\sin x + \cos x}  \\ f(x) =\frac { (\cos x+ \sin x)(\cos  x - \sin  x)}{\sin x + \cos x}  \\ f(x)= \cos x - \sin x \\ f'(x)= -sin x- \cos x \\ f'(\frac {\pi}{2})= -1

Soal 7. $f(x)= \frac { 1- \tan ^2 4x}{\tan 4x} $ Nilai dari $f'(\frac {\pi}{16})=....$
Jawab:
$ f(x)= \frac { 1- \tan ^2 4x}{\tan 4x} \\ f(x)= \frac {1}{\tan 4x} - \frac {\tan ^2 4x}{\tan 4x} = \cot 4x- \tan 4x \\ f'(x)=-4 \csc^2 4x -4 \sec^2 4x \\ f'(\frac {\pi}{16})=-4 \csc ^2 4(\frac {\pi}{16}) -4 \sec^2 4(\frac {\pi}{16}) = -4 (\sqrt 2)^2 - 4(\sqrt2 )^2=-16$

Soal 8. $f(x)= \frac { 1+ \cos x}{\sin x}$ , maka nilai $f'(\frac {\pi}{4}) $=....
Jawab:
$f(x)= \frac { 1+ \cos x}{\sin x} \\ f(x)= \frac {1}{\sin x}+ \frac {\cos x}{\sin x} \\ f(x)= \csc x+ \cot x \\ f'(x)= -\csc x \cot x - \csc ^2 x \\ f'(\frac {\pi}{4}) = - \sqrt 2.1 - 2 = - \sqrt 2 -2$

Soal 9. f(x) = $ \sqrt {\tan ^{-1} x}$ maka nilai dari $ f' ( \frac {3 \pi}{4})=...$
Jawab: $ f(x) =  \sqrt {\tan ^{-1} x} \\ f(x) = \sqrt {\cot x }\\ f(x)= (\cot x)^ { \frac {1}{2}} \\ f'(x) = \frac {1}{2} (\cot x)^{- \frac {1}{2}} \sec ^2 x \\ f'(\frac { 3\pi}{4}) = \frac {1}{2} (\cot (\frac { 3\pi}{4}))^{- \frac {1}{2}} \sec ^2 (\frac { 3\pi}{4})$

Soal 10: $ f(x) =  \frac {3}{sin 5x + cos (90^\circ -3x)} $Nilai dari $f'(\frac {\pi}{4})$
Jawab:
$ f(x) =  \frac {3}{sin 5x + cos (90^\circ -3x)} \\ f(x)= \frac {3}{sin 5x+sin 3x}  \\ u = 3 \rightarrow u'=0 \\ v= sin 5x +sin 3x \rightarrow v'=5 cos 5x+ 3 sin 3x \\  f'(x)= \frac {u'v-uv'}{v^2} \\  f'(x)= \frac {0.(sin 5x +sin 3x)-3(5 cos 5x+ 3 sin 3x)}{(sin 5x +sin 3x )^2} \\ f'(\frac {\pi}{4})= \frac {0-3(- 5.\frac {1}{2}\sqrt2)+3 \frac {1}{2}\sqrt2) }{-\frac {1}{2}\sqrt2+\frac {1}{2}\sqrt2} = \frac {...}{0}= \infty $