为什么lnx的导数是1/x,顺便问问为什么∫1/x=lnx,

Question

为什么ln x的导数是1/x,
顺便问问为什么∫1/x=ln x,

Answer 1

f(x) = lnx
d/dx (lnx),Based on the definition of derviative
= lim(h->0) [f(x+h)-f(x)]/h
= lim(h->0) [ln(x+h)-lnx]/h
= lim(h->0) (1/h)ln[(x+h)/x]
= lim(h->0) (1/h)(x/h)(h/x)ln(1+h/x)
= lim(h->0) (1/x)ln[(1+h/x)^(x/h)]
= (1/x)lim(h->0) ln{(1+h/x)^[1/(h/x)]}
= (1/x) * lne,Definition of e = lim(x->0) (1+x)^(1/x)
= 1/x
∵d/dx (lnx) = 1/x,integrate both sides,we have
∫ d/dx (lnx) dx = ∫ (1/x) dx
∫ d(lnx) = ∫ (1/x) dx
=> ∫ (1/x) dx = lnx + C
When x>1,it is obvious that ∫ (1/x) dx = lnx + C
When x<1,let -y = x,where y>0 and -dy = dx
∫ (1/x) dx = ∫ 1/(-y) (-dy) = ∫ 1/y dy = lny + C = ln(-x) + C for x<0
∴∫ (1/x) dx = ln(±x) + C = ln|x| + C