第1个回答 2018-08-07
1）lim(y→0+)F(x,y)
= lim(y→0+)∫[-y,y]f(x+t)dt/(2y) (0/0)
= lim(y→0+)[f(x+y)-f(x-y)(-1)]/2
= f(x)；
2）DF/Dx
= (D/Dx){∫[-y,y]f(x+t)dt/(2y)}
= ∫[-y,y]f'(x+t)dt/(2y)
= [f(x+y)-f(x-y)]/(2y)；
3）lim(y→0+)(DF/Dx)
= lim(y→0+)[f(x+y)-f(x-y)]/(2y) (0/0)
= lim(y→0+)[f'(x+y)-f'(x-y)(-1)]/2
= f'(x)。追问
= lim(y→0+)∫[-y,y]f(x+t)dt/(2y) (0/0)
= lim(y→0+)[f(x+y)-f(x-y)(-1)]/2
= f(x)；
2）DF/Dx
= (D/Dx){∫[-y,y]f(x+t)dt/(2y)}
= ∫[-y,y]f'(x+t)dt/(2y)
= [f(x+y)-f(x-y)]/(2y)；
3）lim(y→0+)(DF/Dx)
= lim(y→0+)[f(x+y)-f(x-y)]/(2y) (0/0)
= lim(y→0+)[f'(x+y)-f'(x-y)(-1)]/2
= f'(x)。追问
谢谢你^_^
第2个回答 2018-08-06
请问第一小问怎么得出的
追答用的是洛必达法则。
其中：分子求导，用到变限函数求导公式
谢谢
本回答被提问者采纳
相似回答
