Kumpulan Soal Turunan Trigonometri Dan Pembahasan
Di bawah ini terdapat 15 soal turunan trigonometri dalam bentuk essay beserta kunci jawaban. Semoga soal turunan trigonometri di bawah ini bisa menambah kemampuan untuk berlatih dalam rangka mengasah kemampuan bermatematika khusus dalam topik turunan trigonometri.
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,
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 bagian sehingga dipakai rumus turunan untuk fungsi pecahan.
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 ibarat 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 $ Sumber http://www.marthamatika.com/
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 bagian sehingga dipakai 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 ibarat 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 $ Sumber http://www.marthamatika.com/
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