函数F(x)=∫f(2x+1)dx的导数为(1/2)*f(2x+1),对吗?
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你们两个都错。
设G(x)为ƒ(x)的原函数。
F(x)
=
∫
ƒ(2x
+
1)
dx
=
(1/2)∫
ƒ(2x
+
1)
d(2x
+
1)
=
(1/2)G(2x
+
1)
+
C
F'(x)
=
(1/2)G'(2x
+
1)
*
2
=
ƒ(2x
+
1)
或用换元法:令v
=
2x
+
1,dv
=
2
dx
∫
ƒ(2x
+
1)
dx
=
∫
ƒ(v)
*
(1/2)dv
=
(1/2)∫
ƒ(v)
dv
=
(1/2)G(v)
+
C
F'(x)
=
(1/2)
*
G'(v)
*
v'
=
(1/2)ƒ(2x
+
1)
*
2
=
ƒ(2x
+
1)
设G(x)为ƒ(x)的原函数。
F(x)
=
∫
ƒ(2x
+
1)
dx
=
(1/2)∫
ƒ(2x
+
1)
d(2x
+
1)
=
(1/2)G(2x
+
1)
+
C
F'(x)
=
(1/2)G'(2x
+
1)
*
2
=
ƒ(2x
+
1)
或用换元法:令v
=
2x
+
1,dv
=
2
dx
∫
ƒ(2x
+
1)
dx
=
∫
ƒ(v)
*
(1/2)dv
=
(1/2)∫
ƒ(v)
dv
=
(1/2)G(v)
+
C
F'(x)
=
(1/2)
*
G'(v)
*
v'
=
(1/2)ƒ(2x
+
1)
*
2
=
ƒ(2x
+
1)
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