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)}

$\begin{array}{rl}f(x)& ={\sin }^3(3x^2-2)\\ f^{\prime }(x)& =3{\sin }^2(3x^2-2)\cos (3x^2-2)\cdot 6x\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 18x{\sin }^2(3x^2-2)\cos (3x^2-2)}}}}\end{array}$

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

$\begin{array}{rl}y& ={\cos }^{-1}x\\ & ={\displaystyle \frac{1}{\cos x}}\\ & =\sec x\\ y^{\prime }& =\tan x\sec x\end{array}$

$\begin{array}{rl}u& =\tan x\\ u^{\prime }& ={\sec }^{2}x\end{array}$

$\begin{array}{rl}v& =\sec x\\ v^{\prime }& =\tan x\sec x\end{array}$

$\begin{array}{rl}{y}^{\prime }& =u\cdot v\\ y"& =u^{\prime }v+u{v}^{\prime }\\ & ={\sec }^{2}x\sec x+\tan x\tan x\sec x\\ & ={\sec }^{3}x+{\tan }^{2}x\sec x\\ & ={\displaystyle \frac{1}{{\cos }^{3}x}}+{\displaystyle \frac{{\sin }^{2}x}{{\cos }^{2}x}}{\displaystyle \frac{1}{\cos x}}\\ & ={\displaystyle \frac{1}{{\cos }^{3}x}}+{\displaystyle \frac{{\sin }^{2}x}{{\cos }^{3}x}}\\ & ={\displaystyle \frac{1+{\sin }^{2}x}{{\cos }^{3}x}}\\ & ={\displaystyle \frac{1+1-{\cos }^{2}x}{{\cos }^{3}x}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{2-{\cos }^{2}x}{{\cos }^{3}x}}}}}}\end{array}$

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

$\begin{array}{rl}u& =t\\ u^{\prime }& =1\end{array}$

$\begin{array}{rl}v& =\sin t\\ v^{\prime }& =\cos t\end{array}$

$\begin{array}{rl}f(t)& =uv\\ f^{\prime }(t)& =u^{\prime }v+u{v}^{\prime }\\ & =1\cdot \sin t+t\cos t\\ & =\sin t+t\cos t\end{array}$

$\begin{array}{rl}u& =t\\ u^{\prime }& =1\end{array}$

$\begin{array}{rl}v& =\cos t\\ v^{\prime }& =-\sin t\end{array}$

$\begin{array}{rl}f"(t)& =\cos t+1\cdot \cos t+t\cdot (-\sin t)\\ & =\cos t+\cos t-t\sin t\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -t\sin t+2\cos t}}}}\end{array}$

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

$\begin{array}{rl}u& =x\\ u^{\prime }& =1\end{array}$

$\begin{array}{rl}v& =\cos x\\ v^{\prime }& =-\sin x\end{array}$

$\begin{array}{rl}y& =uv\\ y^{\prime }& =u^{\prime }v+u{v}^{\prime }\\ & =1\cdot \cos x+x(-\sin x)\\ & =\cos x-x\sin x\end{array}$

$\begin{array}{rl}u& =x\\ u^{\prime }& =1\end{array}$

$\begin{array}{rl}v& =\sin x\\ v^{\prime }& =\cos x\end{array}$

$\begin{array}{rl}y"& =-\sin x-(1\cdot \sin x+x\cos x)\\ & =-\sin x-\sin x-x\cos x\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -x\cos x-2\sin x}}}}\end{array}$

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

$\begin{array}{rl}y& ={\tan }^2(3x-2)\\ y^{\prime }& =2\tan (3x-2){\sec }^2(3x-2)\cdot 3\\ & =6\tan (3x-2){\sec }^2(3x-2)\end{array}$

$\begin{array}{rl}u& =6\tan (3x-2)\\ u^{\prime }& =6\cdot 3{\sec }^2(3x-2)\\ & =18{\sec }^2(3x-2)\end{array}$

$\begin{array}{rl}v& ={\sec }^2(3x-2)\\ v^{\prime }& =2\sec (3x-2)\cdot 3\tan (3x-2)\sec (3x-2)\\ & =6\tan (3x-2){\sec }^2(3x-2)\end{array}$

$\begin{array}{rl}y"& =u^{\prime }v+u{v}^{\prime }\\ & =18{\sec }^2(3x-2){\sec }^2(3x-2)+6\tan (3x-2)6\tan (3x-2){\sec }^2(3x-2)\\ & =18{\sec }^4(3x-2)+36{\tan }^2(3x-2){\sec }^2(3x-2)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 36{\tan }^2(3x-2){\sec }^2(3x-2)+18{\sec }^4(3x-2)}}}}\end{array}$

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

$\begin{array}{rl}f(x)& =2\sin x+\cos x\\ f^{\prime }(x)& =2\cos x-\sin x\\ f^{\prime }\left({\displaystyle \frac{\pi }{2}}\right)& =2\cos \left({\displaystyle \frac{\pi }{2}}\right)-\sin \left({\displaystyle \frac{\pi }{2}}\right)\\ & =2(0)-1\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -1}}}}\end{array}$

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

$\begin{array}{rl}u& =\sin x\\ u^{\prime }& =\cos x\end{array}$

$\begin{array}{rl}v& =\cos x\\ v^{\prime }& =-\sin x\end{array}$

$\begin{array}{rl}{f}^{\prime }(x)& =u^{\prime }v+u{v}^{\prime }\\ & =\cos x\cos x+\sin x(-\sin x)\\ & ={\cos }^{2}x-{\sin }^{2}x\\ f^{\prime }\left({\displaystyle \frac{\pi }{6}}\right)& ={\cos }^2{\displaystyle \frac{\pi }{6}}-{\sin }^2{\displaystyle \frac{\pi }{6}}\\ & ={\left({\displaystyle \frac{1}{2}}\sqrt{3}\right)}^2-{\left({\displaystyle \frac{1}{2}}\right)}^2\\ & ={\displaystyle \frac{3}{4}}-{\displaystyle \frac{1}{4}}\\ & ={\displaystyle \frac{2}{4}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{1}{2}}}}}}\end{array}$

CARA 2

$\begin{array}{rl}f(x)& =\sin x\cos x\\ & ={\displaystyle \frac{1}{2}}\cdot 2\sin x\cos x\\ & ={\displaystyle \frac{1}{2}}\sin 2x\\ f^{\prime }(x)& ={\displaystyle \frac{1}{2}}\cdot 2\cos 2x\\ & =\cos 2x\\ f^{\prime }\left({\displaystyle \frac{\pi }{6}}\right)& =\cos 2\left({\displaystyle \frac{\pi }{6}}\right)\\ & =\cos {\displaystyle \frac{\pi }{3}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{1}{2}}}}}}\end{array}$

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

$\begin{array}{rl}u& =\sin x+\cos x\\ u^{\prime }& =\cos x-\sin x\end{array}$

$\begin{array}{rl}v& =\cos x\\ v^{\prime }& =-\sin x\end{array}$

$\begin{array}{rl}f(x)& ={\displaystyle \frac{u}{v}}\\ f^{\prime }(x)& ={\displaystyle \frac{u^{\prime }v-u{v}^{\prime }}{v^2}}\\ & ={\displaystyle \frac{(\cos x-\sin x)\cos x-(\sin x+\cos x)(-\sin x)}{(\cos x{)}^2}}\\ & ={\displaystyle \frac{{\cos }^{2}x-\cancel{\sin x\cos x}+{\sin }^{2}x+\cancel{\sin x\cos x}}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{1}{{\cos }^{2}x}}\\ f^{\prime }\left({\displaystyle \frac{\pi }{6}}\right)& ={\displaystyle \frac{1}{{\cos }^2{\displaystyle \frac{\pi }{6}}}}\\ & ={\displaystyle \frac{1}{{\left({\displaystyle \frac{1}{2}}\sqrt{3}\right)}^2}}\\ & ={\displaystyle \frac{1}{{\displaystyle \frac{3}{4}}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{4}{3}}}}}}\end{array}$

CARA 2

$\begin{array}{rl}f(x)& ={\displaystyle \frac{\sin x+\cos x}{\cos x}}\\ & ={\displaystyle \frac{\sin x}{\cos x}}+{\displaystyle \frac{\cos x}{\cos x}}\\ & =\tan x+1\\ f^{\prime }(x)& ={\sec }^{2}x\\ f^{\prime }\left({\displaystyle \frac{\pi }{6}}\right)& ={\sec }^2{\displaystyle \frac{\pi }{6}}\\ & ={\left({\displaystyle \frac{2}{\sqrt{3}}}\right)}^2\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{4}{3}}}}}}\end{array}$

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)
   
   $\begin{array}{rl}u& =x+5\\ u^{\prime }& =1\end{array}$

   $\begin{array}{rl}v& =\sec (2x-3)\\ v^{\prime }& =2\tan (2x-3)\sec (2x-3)\end{array}$
   
   $\begin{array}{rl}{f}^{\prime }(x)& =u^{\prime }v+u{v}^{\prime }\\ & =1\cdot \sec (2x-3)+(x+5)2\tan (2x-3)\sec (2x-3)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle \sec (2x-3)+2(x+5)\tan (2x-3)\sec (2x-3)}}}}\end{array}$


2. g(x)=\sin2x\cos(x-9)
   
   $\begin{array}{rl}u& =\sin 2x\\ u^{\prime }& =2\cos 2x\end{array}$

   $\begin{array}{rl}v& =\cos (x-9)\\ v^{\prime }& =-\sin (x-9)\end{array}$
   
   $\begin{array}{rl}{g}^{\prime }(x)& =u^{\prime }v+u{v}^{\prime }\\ & =2\cos 2x\cos (x-9)+\sin 2x(-\sin (x-9))\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 2\cos 2x\cos (x-9)-\sin 2x\sin (x-9)}}}}\end{array}$

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

$\begin{array}{rl}{f}^{\prime }(x)& =0-2\cdot {\displaystyle \frac{\pi }{2}}\cos {\displaystyle \frac{\pi x}{2}}\\ & =-\pi \cos {\displaystyle \frac{\pi x}{2}}\end{array}$

$\begin{array}{rl}{f}^{\prime }(x)& =0\\ -\pi \cos {\displaystyle \frac{\pi x}{2}}& =0\\ \cos {\displaystyle \frac{\pi x}{2}}& =0\\ \cos {\displaystyle \frac{\pi x}{2}}& =\cos {\displaystyle \frac{\pi }{2}}\end{array}$

$\begin{array}{rll}{\displaystyle \frac{\pi x}{2}}& ={\displaystyle \frac{\pi }{2}}+2k\pi & \times {\displaystyle \frac{2}{\pi }}\\ x& =1+4k\\ x& =1\end{array}$

$\begin{array}{rll}{\displaystyle \frac{\pi x}{2}}& =-{\displaystyle \frac{\pi }{2}}+2k\pi & \times {\displaystyle \frac{2}{\pi }}\\ x& =-1+4k\\ x& =3\end{array}$

x_1=3, x_2=1

$\begin{array}{rl}{x_1}^2+x_2& =3^2+1\\ & =9+1\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 10}}}}\end{array}$

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

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