f(x)=∫(0,2x)f(t/2)dt+ln2,显然f(0)=ln2 两边求导 f'(x)=f(2x/2)*(2x)' 即f'(x)=2f(x)为什么不用将dt配成d(2/t),原式变成f(x)=2∫(0,2x)f(t/2)d2/t)+ln2两边求导f'(x)=2f(2x/2)*(2x)'即f'(x)=4f(x)
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![f(x)=∫(0,2x)f(t/2)dt+ln2,显然f(0)=ln2 两边求导 f'(x)=f(2x/2)*(2x)' 即f'(x)=2f(x)为什么不用将dt配成d(2/t),原式变成f(x)=2∫(0,2x)f(t/2)d2/t)+ln2两边求导f'(x)=2f(2x/2)*(2x)'即f'(x)=4f(x)](/uploads/image/z/13932896-32-6.jpg?t=f%28x%29%3D%E2%88%AB%280%2C2x%29f%28t%2F2%29dt%2Bln2%2C%E6%98%BE%E7%84%B6f%280%29%3Dln2+%E4%B8%A4%E8%BE%B9%E6%B1%82%E5%AF%BC+f%27%28x%29%3Df%282x%2F2%29%2A%282x%29%27+%E5%8D%B3f%27%28x%29%3D2f%28x%29%E4%B8%BA%E4%BB%80%E4%B9%88%E4%B8%8D%E7%94%A8%E5%B0%86dt%E9%85%8D%E6%88%90d%282%2Ft%29%2C%E5%8E%9F%E5%BC%8F%E5%8F%98%E6%88%90f%28x%29%3D2%E2%88%AB%280%2C2x%29f%28t%2F2%29d2%2Ft%29%2Bln2%E4%B8%A4%E8%BE%B9%E6%B1%82%E5%AF%BCf%27%28x%29%3D2f%282x%2F2%29%2A%282x%29%27%E5%8D%B3f%27%28x%29%3D4f%28x%29)
f(x)=∫(0,2x)f(t/2)dt+ln2,显然f(0)=ln2 两边求导 f'(x)=f(2x/2)*(2x)' 即f'(x)=2f(x)为什么不用将dt配成d(2/t),原式变成f(x)=2∫(0,2x)f(t/2)d2/t)+ln2两边求导f'(x)=2f(2x/2)*(2x)'即f'(x)=4f(x)
f(x)=∫(0,2x)f(t/2)dt+ln2,显然f(0)=ln2 两边求导 f'(x)=f(2x/2)*(2x)' 即f'(x)=2f(x)
为什么不用将dt配成d(2/t),原式变成f(x)=2∫(0,2x)f(t/2)d2/t)+ln2
两边求导
f'(x)=2f(2x/2)*(2x)'
即f'(x)=4f(x)
f(x)=∫(0,2x)f(t/2)dt+ln2,显然f(0)=ln2 两边求导 f'(x)=f(2x/2)*(2x)' 即f'(x)=2f(x)为什么不用将dt配成d(2/t),原式变成f(x)=2∫(0,2x)f(t/2)d2/t)+ln2两边求导f'(x)=2f(2x/2)*(2x)'即f'(x)=4f(x)
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)
当把t/2看成整个变量u时,u的积分区间是[0,x],就没有后面的2x求导了.