f(x)=a^x

f'(x)=lim x-->0 /h
(a^x)'=a^x(a^h-1)/h

so ln(x) = lim h-->0 /h or lim h-->00 h

so

ln(x)=

``````
.data
smallllnumber REAL8 0.00000000001 ; how much accurate
x	REAL8 12.0 ; your number
.code
fld smallllnumber
fld x
fyl2x ;x*log2Y
f2xm1 ;Y^x-1
fdiv smallllnumber
``````

result in st(0)

bye

eko
Posted on 2002-11-12 10:27:54 by eko
Nice, but isn't this simpler ?
``````	fld1
fld	x
fyl2x
fldl2e
fdiv``````
Posted on 2002-11-14 04:38:26 by micmic
if f(a) = a^x
f'(a^x)=x(a^(x-1))
Posted on 2002-11-14 17:56:07 by The Svin

if f(a) = a^x
f'(a^x)=x(a^(x-1))

i meant
(a^x)'
you can consider it as f'(x)
Posted on 2002-11-15 15:47:04 by eko
Do you mean that argument of function is x here, and x is logarithm of f(x)=a^x based a?
in other words
if f(x)=a^x
then x = log_a a^x
where a is some constant. ?
so that for example
if a = e
then x =log_e e^x = ln e^x ?
then if delta x here --> 0
then delta function --> 1 since x^0 = 1 if x <> 0
so if f(x)=const^x then 'f(x) =1 with lim_x --> 0
Anyway eko you need transform your
expression: lim h-->00 h
'cause the above means ~ 0.
if h --> 0 then h --> 0. 'Cause x*0=0
You need exclude h in transformations, otherwise the above is useless.
IMHO.

Might be I'm just dummy and you can give some more explanations.
Posted on 2002-11-15 17:34:01 by The Svin
f(x)=a^x ; yes a is a constant.

f'(x) = a^x*ln(a)
if a = e its the most beatifaul function . this is the only function that the derivative , integral and the funtion are equal

now we forget about the x
and pay attention only to a

so ln(a)=lim h-->0 /h
that can be write as
lim h-->00 h ;00 as infinite
Posted on 2002-11-15 19:05:46 by eko