A-level/STPM/Pre-U -Differentiate lnx

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[mathmathician]

First principle:

d/dx(ln(x))
= lim_h->0((ln(x+h)-ln(x))/h)
= lim_h->0((ln(1+h/x))/h)
= lim_h->0(ln((1+h/x)^(1/h))
Given e^a=lim_h->0((1+ah)^(1/h),
where a is a constant.
So,
= ln(e^(1/x))
=1/x

Hence, d/dx(ln(x) = 1/x.

[cesbon]

Ah.. finally some A level material. But here's my another idea almost similar to the way sir did.
ln(x)
Let y=ln(x)
e^y=x
d/dx (e^y) = d/dx (x)
(dy/dy) (d/dx) e^y=1
(dy/dx) (d/dy) e^y=1
(dy/dx) e^y =1
dy/dx = 1/e^y
dy/dx = 1/x
.
.
Happy learning!

[halidi]

I thought we should use the definition to find the derivative 😅