Page 27 - Kelas XII_Matematika Peminatan_KD 3.3

3. Kunci Jawaban Evaluasi.
No. Kunci Keterangan

1. E f(x) = (3x – 5)cos x.
2
′
Misal U(x) = 3x – 5 → ( ) = = 6
2

V(x) = sin x →V’(x)= = − sin

f’(x) = U’(x).V(X) + U(X).V’(x)
f’(x) = 6x.cos x + (3x – 5)(-sin x)
2
f’(x) = 6xcosx –(3x – 5)cos x
2
2. A f(x) = cos³ x
f’(x) = 3.cos x.(-sin x) = -3cos x.sin x = -3cos x.cos x.sin x
2
2
2
2
f’(x) = -3cos x. cos x.sin x (dikalikan 1 = )
2 2
3
3
f’(x) = - .cos x.(2 cos x.sin x) = - cos x sin 2x.
2 2
B ( ) = cos
sin + cos
−sin (sin + cos ) − (cos − sin )cos
′( ) =
(sin + cos ) 2
2
2
= −sin −sin cos −cos +sin cos
3. (sin +cos ) 2
= −1
(sin +cos ) 2

−1 −1 1
′ ( ) = = = −
4 2 1 1 2 2
(sin + cos ) ( √2 + √2)
4 4 2 2
2
4 D ( ) = sin (2 + 3)
′( ) = 2sin(2 + 3) × cos(2 + 3) × 2 = 2sin(4 + 6)
5. A y=3sin 2x – 2cos 3x
= 3. cos 2 . 2 − 2. (− sin 3 ).3=6cos 2x + 6sin 3x

6. E 3 2 2
( ) = √ sin 3 = (sin3 )3
3 √ sin 3 3 2
2
′( ) = 2×3cos3 = 2cos3 = 2cos3 × = 2cos3 √ sin 3
3
3 √sin3 3 √sin3 3 √sin3 3 √ sin 3 sin3
2
3 2
= 2cot3 ⋅ √ sin 3

7. C ( ) = sin (2 + )
2
6

′( ) = 2sin (2 + ) × cos (2 + ) × 2 = 2sin (4 + )
6 6 3

′(0) = 2sin (4 × 0 + ) = 2sin (4 × 0 + ) = √3
3 3
8. B 5
f   x  5 sin x cos x  sin 2 x
2
5 5

f '   x  sin 2 x   2 cos 2 x 5 cos 2 x
2 2

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