Soal 1. f(x)= 3 sin x - 4 cos 5x. f'($ \frac {\pi}{4}$)=...
Jawab:
f'(x)= 3 cos x - 4.5.(-sin 5x) = 3cosx+20sin 5x
f'($ \frac {\pi}{4}$)=3cos $ \frac {\pi}{4}$+20 sin 5.$ \frac {\pi}{4}$
$f'( \frac {\pi}{4})=3 .\frac {1}{2} \sqrt 2 +20 (-\frac {1}{2} \sqrt 2 )= \frac {-17}{2} \sqrt 2$
Soal 2. f(x) = $\sqrt 2$ sin (2x+5)-3 sin (2x+5). f'($ \frac {\pi}{9}$)
Jawab:
$f(x) = \sqrt 2 sin (2x+5)-3 sin (2x+5) \\ f(x) =( \sqrt 2 -3) sin (2x+5) \\ f'(x)=( \sqrt 2 -3). 2. cos (2x+5) \\ f'(\frac {\pi}{9})=( \sqrt 2 -3). 2. cos (2.(\frac {\pi}{9})+5) =2-3 \sqrt2$
Soal 3. 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 4. $f(x)= \frac {\sin 2x + \cos 2x }{\cos 2x} \, \, \, ,f'(\frac {\pi}{6})=...$
Jawab:
$f(x)= \frac {\sin 2x + \cos 2x }{\cos 2x} \\ f(x)= \frac {\sin 2x }{\cos 2x} +1 \\ f(x)= \tan 2x+1 \\ f'(x)= 2.\sec^2 2x \\ f'(\frac {\pi}{6})=2. \sec^2 \frac {\pi}{6} \\f'(\frac {\pi}{6})= 2. (\frac {2}{\sqrt 3})^2 = \frac {8}{3}$
Soal 5. $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 6. $f(x)= \frac {\sin ^2x + \cos ^2x }{\sin ^2x} \, \, \, ,f'(\frac {2 \pi}{3})=... $
Jawab:
$f(x)= \frac {\sin ^2x + \cos ^2x }{\sin ^2x} \\ f(x)= \frac {1}{\sin ^2 x} = \csc^2 x \\ f'(x)= 2 \csc x. (-\csc x . \cot x) \\ f'(x)= -2 \csc x. \csc x . \cot x \\ f'(\frac {2 \pi}{3})=-2 \csc (\frac {2 \pi}{3}). \csc (\frac {2 \pi}{3}) . \cot (\frac {2 \pi}{3}) \\ -2.\frac {2}{\sqrt 3} . \frac {2}{\sqrt 3}. - \frac {1}{\sqrt 3} =\frac {8}{9} \sqrt 3$
Soal 7. 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 8,$f(x)= \frac {6x}{\cos (6x-1)} \, \text {maka } \, f'(x)=...$
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 9. $f(x)= \frac {2}{\sin x \cos x} , \,\, \text {maka } f'(x)=...$
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 10.$f(x)= \frac {\cos 2x}{1- \sin 2x} $ Maka $f (\frac {\pi}{4})=...$
Jawab:
Soal ini berbentuk pecahan, gunakan rumusan seperti nomer 8.
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 11. $f(x)= \frac { \cos 2x}{\sin x + \cos x}$ 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 12. $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 13. $f(x)= \frac { 1+ \cos x}{\sin x}$ , 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 14. 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 15: $ 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 $
Jadilah Komentator Pertama untuk "Kumpulan Soal Turunan Trigonometri dan Pembahasan"
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