Thread: Basic doubt on differential of ln(ax) from first principles

Hi Guys,


I have a fundamental doubt on differential of

y = f(x) = ln(ax)
where,
ln - natual logorithm to base e,
a - constant
x - independent variable.

I need to find the value of dy/dx from first principles.

Can anybody help me on this?

Thanks,
Srini

Derivative ln(x)

Thanks for your reply on derivative of ln(x) friend.
But, I need to know the derivative of ln(ax)
it turns out to be:


(1/x) lim (1+u/3) ^ (1/u)
(u -> 0)

I dont think the above limit result 'e'.
This is where i'm stuck.

let u/3=t then, when u-->0 also t-->0. 1/u=1/3t.

So, (1+u/3) ^ (1/u)=(1+t)^(1/3t)={(1+t)^(t)}^(1/3)=e^{1/3} (when t-->0)

Originally Posted by Also sprach Zarathustra

let u/3=t then, when u-->0 also t-->0. 1/u=1/3t.

So, (1+u/3) ^ (1/u)=(1+t)^(1/3t)={(1+t)^(t)}^(1/3)=e^{1/3} (when t-->0)

Ya, exactly.
Therefore,
derivative of ln(3x) is then, (1/3x) * (3) i.e., (1/x).
Is that right?

Yes, beause ln(3x)= ln(x)+ ln(3) and ln(3) is a constant: the derivative of ln(ax) is exactly the same as the derivative of ln(x) for any positive x.

Yes.(I think...)

But there is better way of solving this...

Recall: ln(a*b)=ln(a)+ln(b)

Now:

ln(ax)=ln(a)+ln(x)

so the derivative of ln(ax) is:

{ln(ax)}'= {ln(a)}' + {ln(x)}'

But ln(a) is a constant! ({ln(a)}'=0)

Hence:

{ln(ax)}'= {ln(a)}' + {ln(x)}'= 0 +1/x=1/x

Thanks so much :-)