Page 11 - Bahan Ajar Berbasis Flipped Learning pada materi Diferensial
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Penyelesaian
3
2
a. ( ) = − − 7
′
2
3
Misal: = − → = −3 3−1 = −3
2
1
′
= 7 → = 7.2. 2−1 = 14 = 14
2
′
′
′
Jadi, jika ( ) = − maka ( ) = − = −3 − 14
b. ( ) = 4 + 15
2
′
2
Misal: = 4 → = 4.2. 2−1 = 8
= 15 → = 15. 1−1 = 15. = 15
0
′
′
′
Jadi, jika ( ) = + maka ( ) = + = 8 + 15
′
3
c. ( ) = 9 − 3 + 5
2
′
2
3
Misal: = 9 → = 9.3. 3−1 = 27
′
′
= 3 → = 3. 1−1 = 3
′
′
= 5 = 5 −2 → = 5. (−2). −2−1 = −10 −3 = −10
2 −3
′
′
′
Jadi, jika ( ) = − + maka ( ) = − +
′
10
′
2
( ) = 27 − 3 −
−3
3
d. ( ) = 5 − √ + 1
2
0
′
Misal: = 5 → = 5. 1−1 = 5 = 5
2
2 2 −1 2 − 1 2
3
= √ = 3 → = 3 = 3 = 3
′
2
3 3 3 √
= 1 → = 0
′
′
Jadi, jika ( ) = − + maka ( ) = − + ′
′
′
′
( ) = 5 − 3 2
3 √
10 | T u r u n a n K e l a s X I S M A
