f(x,n) = x^(n-1) * ln(x)
f'(x,n) = x^(n-1) * (1/x) + (n-1) * x^(n-2) * ln(x)
= x^(n-2) + (n-1) * x^(n-2) * ln(x)
= x^(n-2) + (n-1) * f(x,n-1)
對n做數(shù)學(xué)歸納法.
n = 1時,有
f(x,1) = ln(x),
f'(x,1) = 1/x = 0!/ x.
成立.
設(shè)(n-1)時成立,即
f[n-1階導(dǎo)](x,n-1) = (n-2)!/ x.
則有
f[n階導(dǎo)](x,n)
= ( (x^(n-1) * ln(x))' )[再求n-1階導(dǎo)數(shù)]
= ( x^(n-2) + (n-1) * f(x,n-1) )[求n-1階導(dǎo)數(shù)]
= 0 + (n-1) * f[n-1階導(dǎo)](x,n-1)
= (n-1) * (n-2)!/ x
證畢.