Question

f(x)=∫(0,2x)f(t/2)dt+ln2,显然f(0)=ln2 两边求导 f'(x)=f(2x/2)*(2x)'

Answer 1

如果要d(x/2)的话,注意积分上下限可能有更变的.

将t变为t/2,d(t/2) = (1/2)dt ==> dt = 2 d(t/2)

当t = 0时,t/2 = 0

当t = 2x时,t/2 = 2x/2 = x

所以∫(0→2x) f(t/2) dt = ∫(0→x) f(t/2) * 2 d(t/2) = 2∫(0→x) f(u) du,假设u = t/2

∴f'(x) = 2f(x) * x' = 2f(x)