I thought it would be interesting to derive #d/(dx) (ln x) #

We know the definition of the derivative is:

# (d(f(x)))/dx = lim_(h to 0) (f(x+h)-f(x))/h #

#=> d/(dx) ( lnx) = (ln(x+h) - lnx )/h #

Using our log laws:

#lim_(h to 0) (ln( (x+h)/x ) )/h #

We know # (x+h)/x = 1 + h/x #

#=> lim_(h to 0) (ln(1+h/x))/h #

#color(red)(--------------)#

The not so interesting way:

As #h to 0# the denominator and numirator #-> 0 #

So know we can use the L'Hopitals rule:

#=> lim_(h to 0 ) (d/(dh) ( ln(1+h/x) ))/( d/(dh)( h )) #

#=> lim_(h to 0 ) ((1/x)/(1+h/x) )/1 #

#= 1/x #

#color(red)(--------------)#

Or we could use the initial way:

#=> lim_(h to 0) (1+h/x)^(1/h) = e^(1/x)#

As we know:

#color(blue)(--------------)#

#lim_(phi to oo) (1+ 1/(phi))^phi = e #

#lim_(phi to 0) (1+ phi)^(1/phi) = e #

# therefore lim_(phi to 0) (1+ phi/gamma )^(gamma/phi) = e#

Raising each side to #1/gamma # power

#lim_(phi to 0) (1 + phi/gamma)^(1/phi) = e^(1/gamma) #

#color(blue)(--------------)#

Raising each side by the power of #h#:

#=> lim_(h to 0) (1+h/x) = e^(h/x) #

#=> lim_(h to 0) ln(e^(h/x)) /h #

#=> lim_(h to 0) ((h/x)* ln(e))/h #

#( ln(e) = 1 ) #

#=> lim_(h to 0) 1/x #

#= 1/x #