Question

设F(x)=∫(上限x,下限0)tf(x^2-t^2)dt,则F(x)导数等于

Answer 1

简单计算一下即可，答案如图所示

Answer 2

Answer 3

Answer 4

首先换元,再利用变上限积分求导

令u=x^2 -t^2.则t=√(x^2-u). dt=d√(x^2-u)=1/2•1/√(x^2-u)du
当t=0时,u=x^2.当t=x时,u=0

∴∫(0→x)tf(x^2 -t^2)dt
=∫(x^2→0)√(x^2-u) f(u)d√(x^2-u)
=∫(x^2→0)√(x^2-u) f(u)•(1/2•1/√(x^2-u))du
=∫(x^2→0) f(u)•1/2du
=1/2∫(x^2→0) f(u)du
d/dx•∫(0→x)tf(x^2 -t^2)dt
=d/dx•（1/2∫(x^2→0) f(u)du）
=(1/2∫(x^2→0) f(u)du)′
=1/2•f(x^2)(x^2)′
=1/2•f(x^2)•2x
=xf(x^2)

Answer 5

他的导数过程如下:

追答