proof for clalculus log rules

[brandy]

proof for the differential log rule
proof for the integral log rule

please and thankyou

[CompuChip]

Do you mean why
\frac{d}{dx} {}^a\log(x) = \frac{1}{x \operatorname{ln}(a)}
and
\int {}^a\log(x) \, dx = \frac{1}{\operatorname{ln}(a)}\left( x \log(x) - x \right)?

The second one is easiest: just differentiate the right hand side and check that you get the integrand (alog(x)) back.

For the first one, I assume that you have defined the log function as
a^{{}^a \log(x)} = x
(i.e. it inverts exponentiation) and that you will buy (or have somehow proven) that
\frac{d}{dx} a^x = \operatorname{ln}(a) a^x.

Now consider the derivative of
a^{{}^a\log(x)}
and use the chain rule.

[HallsofIvy]

How you prove such things depends strongly on how you define log x!

It is common, in many calculus texts, to define the natural logarithm by
ln(x)= \int_1^x\frac{1}{t} dt[/itex]

Now the derivative rule follows trivially from the fundamental theorem of calculus. Further, you can show all the properties of the logarithm directly from that definition, including the facts that it is invertible and that its inverse function can be written as a real number to the x power.

Once you have natural logarithm, the same rules and properties follow for log_a x by using
[tex]log_a(x)= \frac{ln(x)}{ln(a)}

Since this has nothing to do with "Differential Equations", I am moving it to "Calculus and Analysis".

[Pinu7]

I think he means the definition from algebra. I also think he means the proof that d(ln x)/dx=1/x.

Am I right? Anyway,

y=ln x\Rightarrowx=e^y

Using implicit differentiation,

(e^y)(dy/dx)=1 so dy/dx=1/(e^y)
but e^y=x so
dy/dx=1/x.

[HallsofIvy]

That works, of course, assuming that you have already proved the derivative formula for e^x. And that involves showing that
\lim_{h\rightarrow 0}\frac{a^h- 1}{h}
exists. I think using the integral definition of ln(x) is much simpler. And I don't see any reason for using an "algebra definition" rather than a "calculus definition" to do a Calculus problem!