Exercise Zone: Turunan Fungsi Trigonometri

Berikut ini adalah kumpulan soal mengenai turunan fungsi trigonometri tipe standar. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

Tipe:

Standar SBMPTN

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No. 1

Turunan pertama dari {f(x)=\sin^3\left(3x^2-2\right)} adalah

1. {2\sin^2\left(3x^2-2\right)\sin\left(6x^2-4\right)}
2. {18x\sin^2\left(3x^2-2\right)\cos\left(3x^2-2\right)}
3. {12\sin^2\left(3x^2-2\right)\cos\left(6x^2-4\right)}

4. {24\sin^3\left(3x^2-2\right)\cos^2\left(3x^2-2\right)}
5. {24\sin^3\left(3x^2-2\right)\cos\left(3x^2-2\right)}

\(\begin{aligned} f(x)&=\sin^3\left(3x^2-2\right)\\ f'(x)&=3\sin^2\left(3x^2-2\right)\cos\left(3x^2-2\right)\cdot 6x\\ &=\boxed{\boxed{18x\sin^2\left(3x^2-2\right)\cos\left(3x^2-2\right)}} \end{aligned}\)

Jadi, turunan pertama dari f(x)=\sin^3\left(3x^2-2\right) adalah 18x\sin^2\left(3x^2-2\right)\cos\left(3x^2-2\right).
JAWAB: B

No. 2

Turunan kedua dari fungsi {y=\cos^{-1}x} adalah y" adalah

1. \dfrac{1-3\cos^2x}{\cos^3x}
2. \dfrac{2+\cos^2x}{\cos^3x}
3. \dfrac{2-\cos^2x}{\cos^3x}

4. \dfrac{2+3\cos^2x}{\cos^3x}
5. \dfrac{2-3\cos^2x}{\cos^3x}

Grup Telegram

\(\eqalign{ y&=\cos^{-1}x\\ &=\dfrac1{\cos x}\\ &=\sec x\\ y'&=\tan x\sec x }\)

\(\eqalign{ u&=\tan x\\ u'&=\sec^2x }\)

\(\eqalign{ v&=\sec x\\ v'&=\tan x\sec x }\)

\(\eqalign{ y'&=u\cdot v\\ y"&=u'v+uv'\\ &=\sec^2x\sec x+\tan x\tan x\sec x\\ &=\sec^3x+\tan^2x\sec x\\ &=\dfrac1{\cos^3x}+\dfrac{\sin^2x}{\cos^2x}\dfrac1{\cos x}\\ &=\dfrac1{\cos^3x}+\dfrac{\sin^2x}{\cos^3x}\\ &=\dfrac{1+\sin^2x}{\cos^3x}\\ &=\dfrac{1+1-\cos^2x}{\cos^3x}\\ &=\boxed{\boxed{\dfrac{2-\cos^2x}{\cos^3x}}} }\)

Jadi, y"=\dfrac{2-\cos^2x}{\cos^3x}.
JAWAB: C

No. 3

Turunan kedua dari fungsi {f(t)=t\sin t} adalah f"(t) adalah ....

1. {t\sin t+2\cos t}
2. {t\sin t-2\cos t}
3. {-2t\sin t+\cos t}

4. {2t\sin t+\cos t}
5. {-t\sin t+2\cos t}

Grup Telegram

\(\eqalign{ u&=t\\ u'&=1 }\)

\(\eqalign{ v&=\sin t\\ v'&=\cos t }\)

\(\eqalign{ f(t)&=uv\\ f'(t)&=u'v+uv'\\ &=1\cdot\sin t+t\cos t\\ &=\sin t+t\cos t }\)

\(\eqalign{ u&=t\\ u'&=1 }\)

\(\eqalign{ v&=\cos t\\ v'&=-\sin t }\)

\(\eqalign{ f"(t)&=\cos t+1\cdot\cos t+t\cdot(-\sin t)\\ &=\cos t+\cos t-t\sin t\\ &=\boxed{\boxed{-t\sin t+2\cos t}} }\)

Jadi, f"(t)=-t\sin t+2\cos t.
JAWAB: E

No. 4

Turunan kedua dari fungsi {y=x\cos x} adalah y" adalah

1. {-x\cos x+2\sin x}
2. {x\cos x-2\sin x}
3. {x\cos x+2\sin x}

4. {-x\cos x-2\sin x}
5. {-x\cos x-\sin x}

Grup Telegram

\(\eqalign{ u&=x\\ u'&=1 }\)

\(\eqalign{ v&=\cos x\\ v'&=-\sin x }\)

\(\eqalign{ y&=uv\\ y'&=u'v+uv'\\ &=1\cdot\cos x+x(-\sin x)\\ &=\cos x-x\sin x }\)

\(\eqalign{ u&=x\\ u'&=1 }\)

\(\eqalign{ v&=\sin x\\ v'&=\cos x }\)

\(\eqalign{ y"&=-\sin x-\left(1\cdot\sin x+x\cos x\right)\\ &=-\sin x-\sin x-x\cos x\\ &=\boxed{\boxed{-x\cos x-2\sin x}} }\)

Jadi, y"=-x\cos x-2\sin x.
JAWAB: D

No. 5

Turunan kedua dari fungsi {y=\tan^2(3x-2)} adalah

1. {54\tan^2(3x-2)\sec^2(3x-2)+18\sec^4(3x-2)}
2. {54\tan^2(3x-2)\sec^2(3x-2)+18\sec^2(3x-2)}
3. {36\tan(3x-2)\sec^2(3x-2)+18\sec^4(3x-2)}

4. {18\tan^2(3x-2)\sec^2(3x-2)+36\sec^4(3x-2)}
5. {18\tan(3x-2)\sec^2(3x-2)+18\sec^4(3x-2)}

Grup Telegram

\(\eqalign{ y&=\tan^2(3x-2)\\ y'&=2\tan(3x-2)\sec^2(3x-2)\cdot3\\ &=6\tan(3x-2)\sec^2(3x-2) }\)

\(\eqalign{ u&=6\tan(3x-2)\\ u'&=6\cdot3\sec^2(3x-2)\\ &=18\sec^2(3x-2) }\)

\(\eqalign{ v&=\sec^2(3x-2)\\ v'&=2\sec(3x-2)\cdot3\tan(3x-2)\sec(3x-2)\\ &=6\tan(3x-2)\sec^2(3x-2) }\)

\(\eqalign{ y"&=u'v+uv'\\ &=18\sec^2(3x-2)\sec^2(3x-2)+6\tan(3x-2)6\tan(3x-2)\sec^2(3x-2)\\ &=18\sec^4(3x-2)+36\tan^2(3x-2)\sec^2(3x-2)\\ &=\boxed{\boxed{36\tan^2(3x-2)\sec^2(3x-2)+18\sec^4(3x-2)}} }\)

Jadi, y"=36\tan^2(3x-2)\sec^2(3x-2)+18\sec^4(3x-2).
TIDAK ADA PILIHAN JAWABAN

No. 6

Jika {f(x)=2\sin x+\cos x}, maka f'\left(\dfrac{\pi}2\right) adalah...

1. -2
2. -1
3. 0

4. 1
5. 2

Grup Telegram

\(\eqalign{ f(x)&=2\sin x+\cos x\\ f'(x)&=2\cos x-\sin x\\ f'\left(\dfrac{\pi}2\right)&=2\cos\left(\dfrac{\pi}2\right)-\sin\left(\dfrac{\pi}2\right)\\ &=2(0)-1\\ &=\boxed{\boxed{-1}} }\)

Jadi, f'\left(\dfrac{\pi}2\right)=-1.
JAWAB: B

No. 7

Diketahui {f(x)=\sin x\cos x}. Nilai turunan f(x) di titik {x=\dfrac{\pi}6} adalah....

1. \dfrac12\sqrt3
2. \dfrac12\sqrt2
3. \dfrac12

4. -\dfrac12
5. -\dfrac12\sqrt2

Grup Telegram

CARA 1

\(\eqalign{ u&=\sin x\\ u'&=\cos x }\)

\(\eqalign{ v&=\cos x\\ v'&=-\sin x }\)

\(\eqalign{ f'(x)&=u'v+uv'\\ &=\cos x\cos x+\sin x(-\sin x)\\ &=\cos^2x-\sin^2x\\ f'\left(\dfrac{\pi}6\right)&=\cos^2\dfrac{\pi}6-\sin^2\dfrac{\pi}6\\ &=\left(\dfrac12\sqrt3\right)^2-\left(\dfrac12\right)^2\\ &=\dfrac34-\dfrac14\\ &=\dfrac24\\ &=\boxed{\boxed{\dfrac12}} }\)

CARA 2

\(\eqalign{ f(x)&=\sin x\cos x\\ &=\dfrac12\cdot2\sin x\cos x\\ &=\dfrac12\sin2x\\ f'(x)&=\dfrac12\cdot2\cos2x\\ &=\cos2x\\ f'\left(\dfrac{\pi}6\right)&=\cos2\left(\dfrac{\pi}6\right)\\ &=\cos\dfrac{\pi}3\\ &=\boxed{\boxed{\dfrac12}} }\)

Jadi, f'\left(\dfrac{\pi}6\right)=\dfrac12.
JAWAB: C

No. 8

Nilai turunan dari {f(x)=\dfrac{\sin x+\cos x}{\cos x}} pada {x=\dfrac{\pi}6} adalah ....

1. \dfrac13
2. \dfrac23
3. 1

4. \dfrac43
5. \dfrac53

Grup Telegram

CARA 1

\(\eqalign{ u&=\sin x+\cos x\\ u'&=\cos x-\sin x }\)

\(\eqalign{ v&=\cos x\\ v'&=-\sin x }\)

\(\eqalign{ f(x)&=\dfrac{u}v\\ f'(x)&=\dfrac{u'v-uv'}{v^2}\\ &=\dfrac{(\cos x-\sin x)\cos x-(\sin x+\cos x)(-\sin x)}{(\cos x)^2}\\ &=\dfrac{\cos^2 x-\cancel{\sin x\cos x}+\sin^2 x+\cancel{\sin x\cos x}}{\cos^2 x}\\ &=\dfrac1{\cos^2 x}\\ f'\left(\dfrac{\pi}6\right)&=\dfrac1{\cos^2\dfrac{\pi}6}\\ &=\dfrac1{\left(\dfrac12\sqrt3\right)^2}\\ &=\dfrac1{\dfrac34}\\ &=\boxed{\boxed{\dfrac43}} }\)

CARA 2

\(\eqalign{ f(x)&=\dfrac{\sin x+\cos x}{\cos x}\\ &=\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\cos x}\\ &=\tan x+1\\ f'(x)&=\sec^2x\\ f'\left(\dfrac{\pi}6\right)&=\sec^2\dfrac{\pi}6\\ &=\left(\dfrac2{\sqrt3}\right)^2\\ &=\boxed{\boxed{\dfrac43}} }\)

Jadi, f'\left(\dfrac{\pi}6\right)=\dfrac43.
JAWAB: D

No. 9

Tentukan turunan pertama fungsi-fungsi berikut.

1. f(x)=(x+5)\sec(2x-3)
2. g(x)=\sin2x\cos(x-9)

Grup

1. f(x)=(x+5)\sec(2x-3)
   
   \(\eqalign{ u&=x+5\\ u'&=1 }\)

   \(\eqalign{ v&=\sec(2x-3)\\ v'&=2\tan(2x-3)\sec(2x-3) }\)
   
   \(\eqalign{ f'(x)&=u'v+uv'\\ &=1\cdot\sec(2x-3)+(x+5)2\tan(2x-3)\sec(2x-3)\\ &=\boxed{\boxed{\sec(2x-3)+2(x+5)\tan(2x-3)\sec(2x-3)}} }\)


2. g(x)=\sin2x\cos(x-9)
   
   \(\eqalign{ u&=\sin2x\\ u'&=2\cos2x }\)

   \(\eqalign{ v&=\cos(x-9)\\ v'&=-\sin(x-9) }\)
   
   \(\eqalign{ g'(x)&=u'v+uv'\\ &=2\cos2x\cos(x-9)+\sin2x\left(-\sin(x-9)\right)\\ &=\boxed{\boxed{2\cos2x\cos(x-9)-\sin2x\sin(x-9)}} }\)

Jadi,

1. f'(x)=\sec(2x-3)+2(x+5)\tan(2x-3)\sec(2x-3)
2. g'(x)=2\cos2x\cos(x-9)-\sin2x\sin(x-9)

No. 10

Turunan fungsi f(x)=2-2\sin\dfrac{\pi x}2 bernilai nol di x_1 dan x_2. Jika 0\leq x\leq4 dan x_1\gt x_2, tentukan nilai {x_1}^2+x_2

Grup

\(\eqalign{ f'(x)&=0-2\cdot\dfrac{\pi}2\cos\dfrac{\pi x}2\\ &=-\pi\cos\dfrac{\pi x}2 }\)

\(\eqalign{ f'(x)&=0\\ -\pi\cos\dfrac{\pi x}2&=0\\ \cos\dfrac{\pi x}2&=0\\ \cos\dfrac{\pi x}2&= \cos\dfrac{\pi}2 }\)

\(\eqalign{ \dfrac{\pi x}2&=\dfrac{\pi}2+2k\pi\ &{\color{red}\times\dfrac2{\pi}}\\ x&=1+4k\\ x&=1 }\)

\(\eqalign{ \dfrac{\pi x}2&=-\dfrac{\pi}2+2k\pi\ &{\color{red}\times\dfrac2{\pi}}\\ x&=-1+4k\\ x&=3 }\)

x_1=3, x_2=1

\(\eqalign{ {x_1}^2+x_2&=3^2+1\\ &=9+1\\ &=\boxed{\boxed{10}} }\)

Jadi, {x_1}^2+x_2=10.

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