Exercise Zone : Turunan Fungsi Trigonometri
Table of Contents
Tipe:
No.
Turunan pertama dari- 2 sin2 (3x2 − 2) sin (6x2 − 4)
- 18x sin2 (3x2 − 2) cos (3x2 − 2)
- 12 sin2 (3x2 − 2) cos (6x2 − 4)
- 24 sin3 (3x2 − 2) cos2 (3x2 − 2)
- 24 sin3 (3x2 − 2) cos (3x2 − 2)
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
f(x)&=\sin^3\left(3x^2-2\right)\\
f'(x)&=3\sin^2\left(3x^2-2\right)\cos\left(3x^2-2\right)\cdot 6x\\
&=\boxed{\boxed{18x\sin^2\left(3x^2-2\right)\cos\left(3x^2-2\right)}}
\end{aligned}\)
Jadi, turunan pertama dari f(x) = sin3 (3x2 − 2) adalah 18x sin2 (3x2 − 2) cos (3x2 − 2).
JAWAB: B
JAWAB: B
No.
Turunan kedua dari fungsi- \(\dfrac{1-3\cos^2x}{\cos^3x}\)
- \(\dfrac{2+\cos^2x}{\cos^3x}\)
- \(\dfrac{2-\cos^2x}{\cos^3x}\)
- \(\dfrac{2+3\cos^2x}{\cos^3x}\)
- \(\dfrac{2-3\cos^2x}{\cos^3x}\)
ALTERNATIF PENYELESAIAN
\(\eqalign{
y&=\cos^{-1}x\\
&=\dfrac1{\cos x}\\
&=\sec x\\
y'&=\tan x\sec x
}\)
\(\eqalign{
y'&=u\cdot v\\
y"&=u'v+uv'\\
&=\sec^2x\sec x+\tan x\tan x\sec x\\
&=\sec^3x+\tan^2x\sec x\\
&=\dfrac1{\cos^3x}+\dfrac{\sin^2x}{\cos^2x}\dfrac1{\cos x}\\
&=\dfrac1{\cos^3x}+\dfrac{\sin^2x}{\cos^3x}\\
&=\dfrac{1+\sin^2x}{\cos^3x}\\
&=\dfrac{1+1-\cos^2x}{\cos^3x}\\
&=\boxed{\boxed{\dfrac{2-\cos^2x}{\cos^3x}}}
}\)
| \(\eqalign{ u&=\tan x\\ u'&=\sec^2x }\) | \(\eqalign{ v&=\sec x\\ v'&=\tan x\sec x }\) |
Jadi, \(y"=\dfrac{2-\cos^2x}{\cos^3x}\).
JAWAB: C
JAWAB: C
No.
Turunan kedua dari fungsi- t sin t + 2 cos t
- t sin t − 2 cos t
- −2t sin t + cos t
- 2t sin t + cos t
- −t sin t + 2 cos t
ALTERNATIF PENYELESAIAN
| \(\eqalign{ u&=t\\ u'&=1 }\) | \(\eqalign{ v&=\sin t\\ v'&=\cos t }\) |
| \(\eqalign{ u&=t\\ u'&=1 }\) | \(\eqalign{ v&=\cos t\\ v'&=-\sin t }\) |
Jadi, f "(t) = −t sin t + 2 cos t .
JAWAB: E
JAWAB: E
No.
Turunan kedua dari fungsi- −x cos x + 2 sin x
- x cos x − 2 sin x
- x cos x + 2 sin x
- −x cos x − 2 sin x
- −x cos x − sin x
ALTERNATIF PENYELESAIAN
| \(\eqalign{ u&=x\\ u'&=1 }\) | \(\eqalign{ v&=\cos x\\ v'&=-\sin x }\) |
| \(\eqalign{ u&=x\\ u'&=1 }\) | \(\eqalign{ v&=\sin x\\ v'&=\cos x }\) |
Jadi, y" = −x cos x − 2 sin x.
JAWAB: D
JAWAB: D
No.
Turunan kedua dari fungsi- 54 tan2 (3x − 2) sec2 (3x − 2) + 18 sec4 (3x − 2)
- 54 tan2 (3x − 2) sec2 (3x − 2) + 18 sec2 (3x − 2)
- 36 tan(3x − 2) sec2 (3x − 2) + 18 sec4 (3x − 2)
- 18 tan2 (3x − 2) sec2 (3x − 2) + 36 sec4 (3x − 2)
- 18 tan(3x − 2) sec2 (3x − 2) + 18 sec4 (3x − 2)
ALTERNATIF PENYELESAIAN
\(\eqalign{
y&=\tan^2(3x-2)\\
y'&=2\tan(3x-2)\sec^2(3x-2)\cdot3\\
&=6\tan(3x-2)\sec^2(3x-2)
}\)
\(\eqalign{
y"&=u'v+uv'\\
&=18\sec^2(3x-2)\sec^2(3x-2)+6\tan(3x-2)6\tan(3x-2)\sec^2(3x-2)\\
&=18\sec^4(3x-2)+36\tan^2(3x-2)\sec^2(3x-2)\\
&=\boxed{\boxed{36\tan^2(3x-2)\sec^2(3x-2)+18\sec^4(3x-2)}}
}\)
| \(\eqalign{ u&=6\tan(3x-2)\\ u'&=6\cdot3\sec^2(3x-2)\\ &=18\sec^2(3x-2) }\) | \(\eqalign{ v&=\sec^2(3x-2)\\ v'&=2\sec(3x-2)\cdot3\tan(3x-2)\sec(3x-2)\\ &=6\tan(3x-2)\sec^2(3x-2) }\) |
Jadi, y" = 36 tan2 (3x − 2) sec2 (3x − 2) + 18 sec4 (3x − 2).
JAWAB: TIDAK ADA PILIHAN JAWABAN
JAWAB: TIDAK ADA PILIHAN JAWABAN
No.
Jika- −2
- −1
- 0
- 1
- 2
ALTERNATIF PENYELESAIAN
\(\eqalign{
f(x)&=2\sin x+\cos x\\
f'(x)&=2\cos x-\sin x\\
f'\left(\dfrac{\pi}2\right)&=2\cos\left(\dfrac{\pi}2\right)-\sin\left(\dfrac{\pi}2\right)\\
&=2(0)-1\\
&=\boxed{\boxed{-1}}
}\)
Jadi, \(f'\left(\dfrac{\pi}2\right)=-1\).
JAWAB: B
JAWAB: B
No.
Diketahui- \(\dfrac12\sqrt3\)
- \(\dfrac12\sqrt2\)
- \(\dfrac12\)
- \(-\dfrac12\)
- \(-\dfrac12\sqrt2\)
ALTERNATIF PENYELESAIAN
CARA 1
| \(\eqalign{ u&=\sin x\\ u'&=\cos x }\) | \(\eqalign{ v&=\cos x\\ v'&=-\sin x }\) |
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, nilai turunan f(x) di titik \(x=\dfrac{\pi}6\) adalah \(\dfrac12\).
JAWAB: C
JAWAB: C
No.
Nilai turunan dari \(f(x)=\dfrac{\sin x+\cos x}{\cos x}\) pada \(x=\dfrac{\pi}6\) adalah ....- \(\dfrac13\)
- \(\dfrac23\)
- 1
- \(\dfrac43\)
- \(\dfrac53\)
ALTERNATIF PENYELESAIAN
CARA 1
| \(\eqalign{ u&=\sin x+\cos x\\ u'&=\cos x-\sin x }\) | \(\eqalign{ v&=\cos x\\ v'&=-\sin x }\) |
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
JAWAB: D
No.
Tentukan turunan pertama fungsi-fungsi berikut.- f(x) = (x + 5) sec (2x − 3)
- g(x) = sin 2x cos (x − 9)
ALTERNATIF PENYELESAIAN
- f(x) = (x + 5) sec (2x − 3)
\(\eqalign{ u&=x+5\\ u'&=1 }\)\(\eqalign{ v&=\sec(2x-3)\\ v'&=2\tan(2x-3)\sec(2x-3) }\)
\(\eqalign{ f'(x)&=u'v+uv'\\ &=1\cdot\sec(2x-3)+(x+5)2\tan(2x-3)\sec(2x-3)\\ &=\boxed{\boxed{\sec(2x-3)+2(x+5)\tan(2x-3)\sec(2x-3)}} }\) - g(x) = sin 2x 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,
- f'(x) = sec (2x − 3) + 2(x + 5) tan (2x − 3) sec (2x − 3)
- g'(x) = 2 cos 2x cos (x − 9) − sin 2x sin (x − 9)
No.
Turunan fungsi \(f(x)=2-2\sin\dfrac{\pi x}2\) bernilai nol diALTERNATIF PENYELESAIAN
\(\eqalign{
f'(x)&=0-2\cdot\dfrac{\pi}2\cos\dfrac{\pi x}2\\
&=-\pi\cos\dfrac{\pi x}2
}\)
\(\eqalign{ f'(x)&=0\\ -\pi\cos\dfrac{\pi x}2&=0\\ \cos\dfrac{\pi x}2&=0\\ \cos\dfrac{\pi x}2&= \cos\dfrac{\pi}2 }\)
x1 = 3 , x2 = 1
\(\eqalign{ {x_1}^2+x_2&=3^2+1\\ &=9+1\\ &=\boxed{\boxed{10}} }\)
\(\eqalign{ f'(x)&=0\\ -\pi\cos\dfrac{\pi x}2&=0\\ \cos\dfrac{\pi x}2&=0\\ \cos\dfrac{\pi x}2&= \cos\dfrac{\pi}2 }\)
| \(\eqalign{ \dfrac{\pi x}2&=\dfrac{\pi}2+2k\pi\ &{\color{red}\times\dfrac2{\pi}}\\ x&=1+4k\\ x&=1 }\) | \(\eqalign{ \dfrac{\pi x}2&=-\dfrac{\pi}2+2k\pi\ &{\color{red}\times\dfrac2{\pi}}\\ x&=-1+4k\\ x&=3 }\) |
\(\eqalign{ {x_1}^2+x_2&=3^2+1\\ &=9+1\\ &=\boxed{\boxed{10}} }\)
Jadi, x12 + x2 = 10.
Post a Comment