My Math Forum Demonstration that the derivative of ln equals 1/x

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 December 7th, 2017, 03:45 PM #1 Newbie   Joined: Dec 2017 From: USA Posts: 2 Thanks: 0 Demonstration that the derivative of ln equals 1/x Hi! Nice to meet you all <333 I would be tremendously grateful if you could possibly give an answer to my questions and help me with this. I have two questions: first of all, what is the limit of an exponential function as x approaches 0? Is it 1? Second one: how exactly could I calculate lim as h approaches 0 of (1+h/x)^(1/h), that somehow must equal e^(1/x)? I am trying to follow the demonstration of why the derivative of ln is the reciprocal function and one step that is unclear is that limit, that is also written as lim of h as it approaches 0 of (1+h)^(1/xh). I tried to make a conjecture using the binomial formula, but I could not find anything (I am honestly not brave to use it for 1/x as an exponent). The only thing I am aware of is that lim of x as it approaches 0 of (1+x)^(1/x)=e. Thank you very much! Hope you can help me. Fondly, ap calc student.
December 8th, 2017, 02:29 AM   #2
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Quote:
 Originally Posted by adalovelace Hi! Nice to meet you all <333 I would be tremendously grateful if you could possibly give an answer to my questions and help me with this. I have two questions: first of all, what is the limit of an exponential function as x approaches 0? Is it 1?
Yes, $\displaystyle e^x$ is defined to be continuous for all x and $\displaystyle e^0= 1$.

Quote:
 Second one: how exactly could I calculate lim as h approaches 0 of (1+h/x)^(1/h), that somehow must equal e^(1/x)? I am trying to follow the demonstration of why the derivative of ln is the reciprocal function and one step that is unclear is that limit, that is also written as lim of h as it approaches 0 of (1+h)^(1/xh). I tried to make a conjecture using the binomial formula, but I could not find anything (I am honestly not brave to use it for 1/x as an exponent). The only thing I am aware of is that lim of x as it approaches 0 of (1+x)^(1/x)=e. Thank you very much! Hope you can help me. Fondly, ap calc student.
How, in your text book, is "e" defined? In many texts, e is defined as the limit, as n goes to infinity, of $\displaystyle \left(1+ \frac{1}{n}\right)^n$. Given that, $\displaystyle e^{1/x}$ would be the limit, as n goes to infinity, of $\displaystyle \left(1+ \frac{1}{n}\right)^{n/x}$. If we let h= 1/n that becomes the limit, as h goes to 0, of $\displaystyle \left(1+ h\right)^{1/xh}$.

Personally, I prefer to do what many modern Calculus texts do: define "ln(x)" by $\displaystyle \ln(x)= \int_1^x \frac{1}{t} dt$. Then it immediately follows, from the 'fundamental theorem of Calculus', that $\displaystyle \frac{d \ln(x)}{dx}= \frac{1}{x}$. Now define $\displaystyle e^x$ to be the inverse function to $\ln(x)$. That is, if $\displaystyle y= \ln(x)$ then $\displaystyle x= e^y$. So $\displaystyle \frac{dy}{dx}= \frac{1}{x}$ becomes $\displaystyle \frac{dx}{dy}= x= e^y$. That is, $\displaystyle \frac{de^y}{dy}= e^y$

Last edited by skipjack; December 8th, 2017 at 10:37 AM.

December 8th, 2017, 03:30 AM   #3
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Quote:
 Originally Posted by Country Boy Yes, $\displaystyle e^x$ is defined to be continuous for all x and $\displaystyle e^0= 1$. How, in your text book, is "e" defined? In many texts, e is defined as the limit, as n goes to infinity, of $\displaystyle \left(1+ \frac{1}{n}\right)^n$. Given that, $\displaystyle e^{1/x}$ would be the limit, as n goes to infinity, of $\displaystyle \left(1+ \frac{1}{n}\right)^{n/x}$. If we let h= 1/n that becomes the limit, as h goes to 0, of $\displaystyle \left(1+ h\right)^{1/xh}$. Personally, I prefer to do what many modern Calculus texts do: define "ln(x)" by $\displaystyle \ln(x)= \int_1^x \frac{1}{t} dt$. Then it immediately follows, from the 'fundamental theorem of Calculus', that $\displaystyle \frac{d \ln(x)}{dx}= \frac{1}{x}$. Now define $\displaystyle e^x$ to be the inverse function to $\ln(x)$. That is, if $\displaystyle y= \ln(x)$ then $\displaystyle x= e^y$. So $\displaystyle \frac{dy}{dx}= \frac{1}{x}$ becomes $\displaystyle \frac{dx}{dy}= x= e^y$. That is, $\displaystyle \frac{de^y}{dy}= e^y$
In fact, when you define the log this way you can prove it satisfies the required properties without mentioning the exponential at all. Namely, define a function $\log: (0,\infty) \to \mathbb{R}$ which satisfies, $\log(ab) = \log(a) + \log(b)$, $\log(\frac{1}{a}) = - \log(a)$ and define $f(x) = \int_1^x \frac{1}{s} \ ds$. We will show that $f$ is the same function.

1. Compute $f(\frac{1}{x})$ using the change of variable $u = \frac{1}{s}$ as
$f(\frac{1}{x}) = \int_1^{\frac{1}{x}} \frac{1}{s} \ ds = -\int_1^x \frac{1}{u} \ du = -f(x)$

2. Compute $f(xy)$ using the change of variable $u = \frac{s}{y}$ combined with the result from 1
$f(xy) = \int_1^{xy} \frac{1}{s} \ ds = \int_{\frac{1}{y}}^x \frac{1}{u} \ du = f(x) - f(\frac{1}{y}) = f(x) + f(y)$

A bit more arguing which I'll omit can show that this is the only such function up to scalar multiples which correspond to the base of the logarithm (i.e. $f(x)$ is exactly $\log(x)$ for some base. Finally, one can solve $f(x) = 1$ using quadrature to obtain
$\int_1^x \frac{1}{s} \ ds = 1 \implies x = \lim_{n\to \infty} \left(1+\frac{1}{n} \right)^n = e$
proving that $f$ is the base $e$ logarithm.

This might not be the route the OP is taking in AP calculus. I would guess the method of derivation there is to define $\ln$ as the inverse of the exponential which means that $e^{\ln(x)} = x$ for all $x$. Then apply the chain rule
$\frac{d}{dx} \left(e^{\ln(x)} \right) = \frac{d \ln(x)}{dx}e^{\ln(x)} = x\frac{d \ln(x)}{dx}$
Taking a derivative on the RHS and dividing $e^{\ln(x)} = x$ gives
$\frac{d \ln(x)}{dx} = \frac{1}{x}$

Last edited by skipjack; December 8th, 2017 at 10:40 AM.

 December 8th, 2017, 04:40 AM #4 Senior Member   Joined: Dec 2015 From: Earth Posts: 238 Thanks: 27 If you can't remember derivative of any function, use inverse function as below: $\displaystyle y'=dy/dx=(dx_{y}/dy)^{-1}=1/x'_{y}$ $\displaystyle (\ln x)'=1/(e^{y})'=1/e^{y}=1/x$ Last edited by skipjack; December 8th, 2017 at 10:14 AM.
 December 8th, 2017, 06:38 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,355 Thanks: 2469 Math Focus: Mainly analysis and algebra Given $$\lim_{x \to 0} (1+x)^{\frac1x} = e$$ we determine $$\lim_{x \to 0} (1+ax)^{\frac1x} = e^a$$ by writing \begin{align*}\lim_{x \to 0} (1+ax)^{\frac1x} &= \lim_{x \to 0} (1+ax)^{\frac{a}{ax}} \\ &= \lim_{x \to 0} \left((1+ax)^{\frac{1}{ax}}\right)^a\end{align*} for sufficiently small $x$, we have $1+ax \gt 0$ and so the expression is continuous and we can write \begin{align*}&& \lim_{x \to 0} (1+ax)^{\frac1x} &= \left( \lim_{x \to 0} (1+ax)^{\frac{1}{ax}}\right)^a \\ &\text{x \to 0 \iff ax \to 0 so} &&= \left( \lim_{ax \to 0} (1+ax)^{\frac{1}{ax}}\right)^a \\ &\text{writing t=ax} &&= \left( \lim_{t \to 0} (1+t)^{\frac{1}{t}}\right)^a \\ &&&= (e)^a = e^a\end{align*} In your case, you have a variable $h$ and a constant $x$. Last edited by v8archie; December 8th, 2017 at 06:41 AM.
 December 8th, 2017, 10:46 AM #6 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,431 Thanks: 105 $\displaystyle y=\log_{a}x\\ \frac{\Delta y}{\Delta x} = \frac{1}{\Delta x}\log_{a}(1+\frac{\Delta x}{x})=\frac{1}{x}\log_{a}(1+\frac{\Delta x}{x})^{\frac{x}{\Delta x}}\\ \lim_{k\rightarrow 0}(1+k)^\frac{1}{k}= e\\ \frac{dy}{dx} = \frac{1}{x}\log_{a}e\\$ or $\displaystyle e^{\log x}=x\\ e^{\log x}\frac{d\log x}{dx}=1$ Last edited by skipjack; December 8th, 2017 at 12:17 PM.
December 21st, 2017, 06:19 AM   #7
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Quote:
 Originally Posted by idontknow If you can't remember derivative of any function, use inverse function as below: $\displaystyle y'=dy/dx=(dx_{y}/dy)^{-1}=1/x'_{y}$ $\displaystyle (\ln x)'=1/(e^{y})'=1/e^{y}=1/x$
What if you can't remember the derivative of the function or its inverse?

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