How to calculate the derivative of ln(x) using the limit

[Cristopher]

Hi.

I know that the derivative of ln(x) is 1/x but I can't seem to find a way to calculate this using this limit

^{lim}_{h→0}\frac{ln(x+h)-ln(x)}{h}

Could anyone give me a hand on how to find derivative of lnx? :)

Thanks!

 

[lugita15]

Welcome to PhysicsForums, Cristopher! Here's a starting point: use the properties of logarithms to turn the whole expression into a single logarithm.

 

[mathwonk]

or use the integral definition of the log.

 

[DonAntonio]

Hi.

I know that the derivative of ln(x) is 1/x but I can't seem to find a way to calculate this using this limit

^{lim}_{h→0}\frac{ln(x+h)-ln(x)}{h}

Could anyone give me a hand? :)

Thanks!

Logarithms are many times studied after the exponential functions, so we can begin with

\frac{\log(x+h)-\log x}{h}=\frac{1}{h}\log\left(1+\frac{h}{x}\right)= \log \left(1+\frac{h}{x}\right)^{1/h}= \log \left[\left(1+\frac{1}{x/h}\right)^{x/h}\right]^{1/x}

and now use that
\left(1+\frac{1}{f(x)}\right)^{f(x)}\xrightarrow [f(x)\to\infty]{} e\,\,,\,\text{whenever}\,\,f(x)\xrightarrow [x\to\infty]{} \infty
and the continuity of both \,\log x\,,\,e^x ...oh, and of course: that these two functions are inverse to each other.

DonAntonio

 

[Cristopher]

Thank you all for your answers!

Welcome to PhysicsForums, Cristopher!

Thank you ;)

Here's a starting point: use the properties of logarithms to turn the whole expression into a single logarithm.

I knew I would probably have to use those properties but I never thought of writing 1/h as an exponent! DonAntonio's post made me realize that I can actually make the indeterminate form 0/0 into 1^∞. The properties ln(a)-ln(b)= ln(a/b) and ln(a^b)= b*ln(a) along with the fact that \displaystyle\lim_{x \to 0}(1+x)^{1/x}=e make this limit quite easy to evaluate:

\displaystyle\lim_{h \to 0}\frac{\ln(x+h)-\ln(x)}{h}=\lim_{h \to 0}\ln \left(\left(\frac{x+h}{x}\right)^{1/h}\right)=\lim_{h \to 0}\ln \left[\left(\left(1+\frac{h}{x}\right)^{x/h}\right)^{h/x*1/h}\right]=\lim_{h \to 0}\ln \left[\left(\left(1+\frac{h}{x}\right)^{x/h}\right)<br /> ^{1/x}\right]=\ln(e^{1/x})=\frac{1}{x}\ln(e)=\frac{1}{x}

This is what you meant, right? lugita15?

or use the integral definition of the log.

I'd never heard about that before. Do you mean this?

\displaystyle\ln(t)=\int^{t}_{1}\frac{1}{x}dx

 

[lugita15]

The properties ln(a)-ln(b)= ln(a/b) and ln(a^b)= b*ln(a) along with the fact that \displaystyle\lim_{x \to 0}(1+x)^{1/x}=e make this limit quite easy to evaluate:

\displaystyle\lim_{h \to 0}\frac{\ln(x+h)-\ln(x)}{h}=\lim_{h \to 0}\ln \left(\left(\frac{x+h}{x}\right)^{1/h}\right)=\lim_{h \to 0}\ln \left[\left(\left(1+\frac{h}{x}\right)^{x/h}\right)^{h/x*1/h}\right]=\lim_{h \to 0}\ln \left[\left(\left(1+\frac{h}{x}\right)^{x/h}\right)<br /> ^{1/x}\right]=\ln(e^{1/x})=\frac{1}{x}\ln(e)=\frac{1}{x}

This is what you meant, right? lugita15?

Yes, this is exactly what I meant. Good job!

I'd never heard about that before. Do you mean this?

\displaystyle\ln(t)=\int^{t}_{1}\frac{1}{x}dx

Yes, that is what mathwonk was referring to. Of course, using that definition makes the problem trivial.

 

[lavinia]

Hi.

I know that the derivative of ln(x) is 1/x but I can't seem to find a way to calculate this using this limit

^{lim}_{h→0}\frac{ln(x+h)-ln(x)}{h}

Could anyone give me a hand? :)

Thanks!

what definition of log are you using?

 

[Cristopher]

what definition of log are you using?

I'm not sure if I understand your question correctly, but I was referring to the natural logarithm, with base e.

 

[HallsofIvy]

That wasn't the question. There are many different ways to define just that function. One was, as noted before, to define
ln(t)= \int_1^t\frac{1}{x}dx

Since 1/x is defined for all non-zero x, that defined ln(t) for all positive t. By the fundamental theorem of Calculus, the derivative of ln(t) is 1/t. And that's positive for positive x so ln(x) is an increasing function. You can then use substitution in the integral to prove that ln(1/t)= -ln(t), ln(st)= ln(s)+ ln(t), and that ln(x^y)= yln(x)

By the mean value theorem, we can write (ln(2)- ln(1))/(2- 1)= 1/c for some c between1 and 2. Of course, that says that ln(2)= 1/c for c< 2 and so ln(2)> 1/2. That's important because then, for any X> 0, ln(2^{2X})= 2Xln(2)&gt; X. Since X could be any number, ln(x) is unbounded. Since ln(t) is increasing, the limit, as x goes to infinity, of ln(x) is infinity and, because ln(1/t) - ln(t), the limit as x goes to 0 is negative infinity. That is, ln(t) maps the set of all positive numbers one-to-one and onto the set of all real numbers. That means it has a inverse function that maps the set of all real numbers to the set of all positive numbers. We can call that inverse function, "exp(x)".

And, finally, here is the important point- If y= exp(x) then x= ln(y). If x is not 0, we can divide both sides by x: 1= (1/x)ln(y)= ln(y^{1/x}). Going back to the exponental form, that says exp(1)= y^{1/x} so that y= (exp(1))^x. That is, the inverse function to ln(x), as defined with that integral, really is an exponential! And, of course, if we define e= ln(1), it follows that [/itex]exp(x)= e^x[/itex].

 

[Millennial]

HallsOfIvy, I think you made a typo. We can't define e as ln(1). We define e as the real solution to the equation ln(x) = 1.

 

[Cristopher]

Lavinia and HallsofIvy, forgive my ignorance, but I didn't know ln(x) could be definied in many ways. If the definition I'm using is so important, I don't know why I haven't been asked this before.

I was taught that ln(x) is a function so that \displaystyle\ f: (0, \infty) \rightarrow \mathbb R That is, the function is defined \displaystyle\forall\ x\in(0, \infty). And, of course, \displaystyle\lim_{x \to +\infty}\ln(x)= +\infty and \displaystyle\lim_{x \to 0^{+}}\ln(x)= -\infty (i.e. ln(x) has a vertical aysmptote at x=0 from the right). I already know that if f(x)=\ln(x) then f^{-1}(x)=e^x because f \circ f^{-1}=x and f^{-1} \circ f=x

But I want to calculate the derivative of ln(x) using the notion of limit. Because I imagine the derivative of ln(x) was calculated for the first time using the definition of the derivative, wasn't it? and I want to find out how they did it. One way, as I have recently discovered thanks to the help of the users here, is to transform the indeterminate form 0/0 into 1^∞, by using the properties of logarithms and some algebraic manipulation.

Does the way I can evaluate this limit depend on the definition of ln(x) I'm using?

\displaystyle\lim_{h \to 0}\frac{\ln(x+h)-\ln(x)}{h}

Here we have the indeterminate form 0/0. So, the question is: Is there any other algebraic manipulation I can do in onder to make this into a determinate form?

By the way, Milennial is right, since ln(1) = 0

 

[Millennial]

The natural logarithm function was invented before the notion of the logarithm itself (or that's what I remember.) It was defined to be the integral of 1/x, because it was not expressible in elementary terms then. It was only later when the notion of the logarithm was invented, the equation log(ab)=log(a)+log(b) became known as the fundamental property of logarithms. Using a simple integral transformation, it is easy to prove this for the integral of 1/x:

\log(ab)=\int_{1}^{ab}\frac{dx}{x}=\int_{1/b}^{a}\frac{dt}{t}=\int_{1/b}^{1}\frac{dt}{t}+\int_{1}^{a}\frac{dt}{t}=\int_{1}^{b}\frac{dt}{t}+\int_{1}^{a}\frac{dt}{t}=\log(b)+\log(a)

This reveals that the function we defined is a logarithm.

Your method requires we defined the exponential function beforehand and also defined the natural logarithm as its inverse. Then, your limit is equal to the limit
\lim_{h\to 0}\log \left( \left( \frac{x+h}{x} \right)^{1/h} \right)
Try to simplify that.