上页 下页返回公式 [u(x)?v(x)]?=u?(x)?v(x)+u(x)?v?(x)的证明:
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l i m
h h
xvxuhxvhxu )()()()(?++
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h h
1 [ u ( x + h ) v ( x + h )? u ( x ) v ( x + h ) + u ( x ) v ( x + h )? u ( x ) v ( x )]
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l i m
h
+
h
xuhxu )()(?v ( x + h ) + u ( x )?
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h
xvhxv )()(
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l i m
h h
xuhxu )()(?+?
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h
v ( x + h ) + u ( x )?
0
l i m
h h
xvhxv )()(?+
=u?(x)v(x)+u(x)v?(x),
[u(x)?v(x)]?
其中
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l i m
h
v ( x + h ) = v ( x ) 是由于 v? ( x ) 存在 。
返回
=
0
l i m
h h
xvxuhxvhxu )()()()(?++
=
0
l i m
h h
1 [ u ( x + h ) v ( x + h )? u ( x ) v ( x + h ) + u ( x ) v ( x + h )? u ( x ) v ( x )]
=
0
l i m
h
+
h
xuhxu )()(?v ( x + h ) + u ( x )?
+
h
xvhxv )()(
=
0
l i m
h h
xuhxu )()(?+?
0
l i m
h
v ( x + h ) + u ( x )?
0
l i m
h h
xvhxv )()(?+
=u?(x)v(x)+u(x)v?(x),
[u(x)?v(x)]?
其中
0
l i m
h
v ( x + h ) = v ( x ) 是由于 v? ( x ) 存在 。
返回