Page 17 - e-book Calculcs I

เฉลย

I. จงหา โดยการใช้สูตร

2
1. ( ) =
( + ℎ) − ( )
′
( ) = lim
ℎ→0 ℎ
2
2
( + 2 ℎ + ℎ ) − 2
= lim
ℎ→0 ℎ
ℎ(2 + ℎ)
= lim
ℎ→0 ℎ
= lim(2 + ℎ) = 2
ℎ→0

2. ( ) = √

( + ℎ) − ( )
′
( ) = lim
ℎ→0 ℎ
√ + ℎ − √
= lim
ℎ→0 ℎ
(√ + ℎ − √ )(√ + ℎ + √ )
= lim
ℎ→0 ℎ(√ + ℎ + √ )
1 1
= lim =
ℎ→0 √ + ℎ + √ 2√

3
3. ( ) = −
3
( + ℎ) = ( + ℎ) − ( + ℎ)
3
2
3
2
= + 3 ℎ + 3 ℎ + ℎ − − ℎ
3
3
2
2
3
( + ℎ) − ( ) = + 3 ℎ + 3 ℎ + ℎ − − ℎ − ( − )
3
3
2
2
3
= + 3 ℎ + 3 ℎ + ℎ − − ℎ − +
2
3
2
= 3 ℎ + 3 ℎ + ℎ − ℎ
2
2
= (3 + 3 ℎ + ℎ − 1)ℎ
จะได้
′
( ) = lim ℎ→0 ( +ℎ)− ( )
ℎ
2
2
(3 +3 ℎ+ℎ −1)ℎ
= lim ℎ→0 ℎ
2
2
= lim ℎ→0 (3 + 3 ℎ + ℎ − 1)
2
= 3 − 1
เพราะฉะนั้น [ − ] = 3 − 1
3
2

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