How do you find the derivative of #ln x#?

2 Answers
Apr 18, 2015

#1/x#

Consider f(x) = lnx and using the first principles work out #f '(x)= lim h->0 {f(x+h)-f(x)}/h#

= #lim h->0 {ln(x+h) -lnx}/h#

=#limh->0{ln((x+h)/x)}/h#

= #lim h->0{ln(1+h/x)}/h#

= #limh->0 1/x {ln(1+h/x)}/ (h/x)#

= #1/x# #limh->0 {ln(1+h/x)}/(h/x)#

= #1/x#, because #[limh->0{ln(1+h/x)}/(h/x) =1]#

Apr 18, 2015

The "how" depends on the definition you are using for #lnx#.

I like: # lnx = int_1^x 1/t dt# and the result is immediate useing the Fundamental Theorem of Calculus.

If you start with a definition of #e^x# and find its derivative, then you probably defined #lnx# by:

#y = lnx# # iff # #x = e^y#

Differentiate implicitely:

#d/(dx)(x) = d/(dx)(e^y)#

#1 = e^y dy/dx#

So
#dy/dx = 1/ e^y = 1/x#