Question

设f(x)连续，Y=∫0~X tf（x^2-t^2）dt 则dy/dx=?

Answer 1

简单计算一下即可，答案如图所示

Answer 2

y
=
∫[0，x]
t
f(x²
-
t²)
dt
令u
=
x²
-
t²，du
=
-2t
dt
当t
=
0，u
=
x²；当t
=
x，u
=
0
y
=
∫[x²，0]
t
f(u)
*
du/(-2t)
=
1/2
∫[0，x²]
f(u)
du
dy/dx
=
1/2
[2x
*
f(x²)
-
0]
=
x
f(x²)
-------------------------------------------------------------------------------------
楼上的方法也不错。
令z²
=
x²
-
t²，2z
dz
=
-2t
dt
=>
dt
=
-z/t
dz
当t
=
0，z
=
x；当t
=
x，z
=
0
∫[0，x]
t
f(x²
-
t²)
dt
=
∫[x，0]
t
f(z²)
*
(-z/t)
dz
=
∫[x，0]
-z
f(z²)
dz
=
∫[0，x]
z
f(z²)
dz
dy/dx
=
x
f(x²)
-
0
=
x
f(x²)
-------------------------------------------------------------------------------------
两个方法也行，但楼上的替换怎么又会涉及因变量y呢？