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Why d/dx(lnx)=1/x?(by definition of derivative)
Let f(x)=lnx
∴ f'(x)=lim h → 0 f ( x + h ) − f ( x )/h
∴ f'(x)=lim h→0 ln(x+h)-lnx/h
=lim h→0 ln(x+h/x)/h
=lim h→0 1/h ln(1+h/x)
=lim h→0 ln(1+h/x)^1/h
Now, ln(lim h→0 (1+h/x)^1/h
NOTE: lim t→∞ (1+1/t)^t=e
put 1/t=u
⇒ t=1/u
∴ as t→∞, u→0
therefore lim u→0 (1+u)^1/u=e
⇒ put h/x=u
as h→0,u→0
therefore ln(lim u→0(1+u)^1/ux
ln(lim u→0{(1+u)^1/u}^1/x
ln e^1/x
1/x
therefore d/dx(lnx)=1/x.
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