Page 49 - e-book Calculcs I

II. จงหาอนุพันธ์ของฟังก์ชัน

1
2
1. ( ) = ∫

1
2
′( ) = [∫ ]

2
= [− ∫ ]
1

2
= − [∫ ]
1
2
= −

2
2. ( ) = ∫ (3 − 1)
1 2

′( ) = [∫ (3 − 1) ]
1
2
2
= (3( ) − 1) ∙ [ ]

2
= (3 − 1)(2 )
3
= 6 − 2

2+3
3. ( ) = ∫ 1 ⁄ 3
3 1+ ⁄ 2
2+3
′( ) = [∫ 1 ⁄ 3 ]
3 1+ ⁄ 2
= ( 1 3 ) [2 + 3 ] − ( 1 3 ) [ 1 ⁄ 3]
1
1+(2+3 ) ⁄ 2 1+( ⁄ 3) ⁄ 2

= 3 3 − ( 1 1 ) ( 1 2 )
1+(2+3 ) ⁄ 2 1+ ⁄ 2 3 ⁄ 3

= 3 3 − 2 1 1
1+(2+3 ) ⁄ 2 3 ⁄ 3(1+ ⁄ 2)

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