Question

proof derivative of log f(x) .. at base a.

Answer 1

Let f(x) = log(x) , with base a not written for convenience..

(f(x+h)-f(x))
f'(x) = lim ---------------
h->0 h

(log(x+h)-log(x))
<=> f'(x) = lim -------------------
h->0 h

log( (x+h)/x )
<=> f'(x) = lim -------------------
h->0 h

1
<=> f'(x) = lim --- . log( (x+h)/x )
h->0 h

<=> f'(x) = lim log( (x+h)/x )1/h
h->0

<=> f'(x) = lim log( (x+h)/x )1/h
h->0

<=> f'(x) = lim log(1 + h/x)1/h
h->0

<=> f'(x) = lim log((1 + h/x)x/h )1/x
h->0

<=> f'(x) = lim (1/x).log(1 + h/x)x/h
h->0

<=> f'(x) =(1/x). lim log(1 + h/x)x/h
h->0

<=> f'(x) =(1/x). lim log(1 + h/x)x/h
h/x->0


<=> f'(x) =(1/x).log lim (1 + h/x)x/h
h/x->0


<=> f'(x) =(1/x).log(e)


<=> f'(x) =(1/x).ln(e)/ln(a)

<=> f'(x) =(1/x)/ln(a)

1
<=> f'(x) = ----------
x. ln(a)

is that what you wanted? see
http://home.scarlet.be/math/exp.htm#Differentiation-of-l

Answer 2

Anyway what you want to do, in words. Is the very similar proof of the ln, you use the definition of derivative and take the limit for h->0. The use log's properties, apply Napier's number, and see what happens.