Question

I am confused is the derivative of 1/x, lnx, or the antiderivative of 1/x, lnx?

Answer 1

Antiderivative of 1/x is ln|x|

Answer 2

Well, +C, but anyway...

Answer 3

If you know why the derivative of e^x is just e^x, then (if you want) we can go through why the derivative of ln(x) is 1/x, and thus, why the antiderivative of 1/x is ln|x|+C.

Answer 4

@lili18 if I remember correctly the relationship is like this

\[\int\limits \frac{ dx }{ x } = \ln(x)+C , \frac{ d }{ dx }\ln(x) = \frac{ 1 }{ x }\]

Answer 5

The derivative of \[\frac{ 1 }{ x } = \ln(x)\] Pretty sure "Anti" in derivative means what's the integral of your function you're talking about.

Answer 6

Photon336, actually,

\(\color{#000000}{ \displaystyle \int\frac{1}{x}dx=\ln|x|+C }\)

With the absolute value.

Answer 7

The absolute value comes from the fact that if x is negative then you get a negative in the logarithm, but with absolute value you will not.

Answer 8

But, bsically, yes, it is the natural log (+c).

Answer 9

I see, that always seemed strange to me, is that just a designation that x can't be negative?

Answer 10

Well, you can't take a log of negative number.
Well, at least, with a positive base, that shouldn't be possible.

Answer 11

Consider, \(\color{#000000}{ \displaystyle \log_a(-b)=c }\)
where, \(\color{#000000}{ \displaystyle a,b>0 }\) (and \(\color{#000000}{ \displaystyle c }\) is just any real)

Then, \(\color{#000000}{ \displaystyle a^c =-b }\) is an impossible case.

Answer 12

Proof for \(\color{#000000}{ \displaystyle d/dx~ (\ln x ) }\), given that \(\color{#000000}{ \displaystyle ~ d/dx~ (e^x ) =e^x }\).

\(\color{#000000}{ \displaystyle y=\ln x }\)
\(\color{#000000}{ \displaystyle e^y=x }\)

(then differentiate both sides, and apply the chain rule of y' to the left side)
\(\color{#000000}{ \displaystyle y'\cdot e^y=1 }\)
\(\color{#000000}{ \displaystyle y=1/ e^y }\)
\(\color{#000000}{ \displaystyle y=1/ x }\)

Answer 13

And thence the integral would follow...

Answer 14

thank you (:

Answer 15

Why " (: " ?

Answer 16

Sorry was just trying to follow through with this myself.

So essentially you're getting this right?

\[e^{y} = x, y = \ln(x)\]


\[e^{\ln(x)} = x\]

\[\frac{ d }{ dx }(e^{\ln(x)}) = \frac{ d }{ dx }*x\]

\[e^{(u)}*\frac{ du }{ dx }, u = \ln(x), \frac{ du }{ dx } = (\frac{ 1 }{ x })\]

\[\frac{ d }{ dx }*(x) = 1\]

\[e^{\ln(x)}*(\frac{ 1 }{ x }) = 1\]

Answer 17

Perhaps your notations are a little off, and I missfollowed, but didn't you just use the hypothesis to prove the hypothesis?

Answer 18

You used d/dx ln(x)=1/x in the middle of your proof. (you can use H to prove H)

Answer 19

tried doing chain rule


\[\frac{ d }{ dx}e^{y} = e^{y}*(\frac{ dy }{ dx}) = e^{y}*(\frac{ 1 }{ x }) = 1\]

Answer 20

And that third step you are exactly applying that d/dx(ln x)=1/x, without having yet proved it.

Answer 21

(I wrote the prove above, in case)

Answer 22

got up to the third line in your proof.

Answer 23

y = 1/x

okay so you substituted

x = e^(y) into

\[\frac{ d }{ dx }\frac{ 1 }{ e^{y} } = x\]

then took the derivatigve of both sides and found that you get 1 on both sides.

\[(d/dx)\frac{ 1 }{ e^{y} }*e^{y} = \frac{ d }{ dx }*x =1, \because (\frac{ 1 }{ e^{y} })*e^{y} =1 \]

Answer 24

I am starting with the following function (because I want to find the derivative of the natural log, which would be the y').
\(\color{#000000}{ \displaystyle y=\ln x }\)

Then, I applied the exponential relationship. That is that,
\(\color{#000000}{ \displaystyle \log_a(b)=c\quad \Longleftrightarrow \quad a^c=b }\)

In this case, I have:
\(\color{#000000}{ \displaystyle e^y=x }\)

Then differentiate both sides.
I have to apply the chain rule (y', because y is a function of x),
to e^y. And, the derivative of x, is just 1. So I get:

\(\color{#000000}{ \displaystyle y'\cdot e^y=1 }\)

next I am just solving for y':
\(\color{#000000}{ \displaystyle y=1/ e^y }\)

And then, I know that \(e^y=x\) from the beginning,
so, all I need to do is to paste that in here.
\(\color{#000000}{ \displaystyle y=1/ x }\)

(Complete)

Answer 25

I was almost done typing the prove using the first principles, but it is ridicuously long when comparing to my relevantly straight-forward proof.

Answer 26

Actually, I will go ahead and post it.
But you can choose to ignore it.
(In fact, if I wasn't me, then I probably would)

Applying first principles to prove the derivative of ln(x).

\(\color{#000000}{ \displaystyle \frac{d}{dx}(\ln x)= \lim_{h\to 0}\frac{\ln(x+h)-\ln(x)}{h} }\)

\(\color{#000000}{ \displaystyle \frac{d}{dx}(\ln x)= \lim_{h\to 0}\frac{1}{h}\ln\left(\frac{x+h}{x}\right) }\)

\(\color{#000000}{ \displaystyle \frac{d}{dx}(\ln x)= \lim_{h\to 0}\ln\left\{\left(\frac{x+h}{x}\right)^{1/h}\right\} }\)

\(\color{#000000}{ \displaystyle \frac{d}{dx}(\ln x)= \lim_{h\to 0}\ln\left\{\left(\frac{x+h}{x}\right)^{x/h}\right\}^{1/x} }\)

\(\color{#000000}{ \displaystyle \frac{d}{dx}(\ln x)=\frac{1}{x} \lim_{h\to 0}\ln\left\{\left(\frac{x+h}{x}\right)^{x/h}\right\} }\)

The above is just algebraic manipulations.
They are easy to understand knowing that
\(c\log_a(b)=\log_a(b^c)\), and the basic
limit properties.

Next, we would evaluate
\(\color{#000000}{ \displaystyle \lim_{h\to 0}\ln\left\{\left(\frac{x+h}{x}\right)^{x/h}\right\}=\lim_{h\to 0}\ln\left\{\left(1+\frac{h}{x}\right)^{x/h}\right\} = }\)

\(\color{#000000}{ \displaystyle=\ln\left\{ \lim_{h\to 0}\left(1+\frac{h}{x}\right)^{x/h}\right\} }\)

Using, \(e^{\ln y}=y\), I can write, (Note: exp(y) is e^y),

\(\color{#000000}{ \displaystyle =\ln \left\{ \lim_{h\to 0}\exp\left\{\ln\left(1+\frac{h}{x}\right)^{x/h}\right\} \right\} }\)

\(\color{#000000}{ \displaystyle =\ln \left\{ \lim_{h\to 0}\exp\left\{\frac{x}{h}\ln\left(1+\frac{h}{x}\right)\right\} \right\} }\)

\(\color{#000000}{ \displaystyle =\ln \left\{ \lim_{h\to 0}\exp\left\{\frac{\ln\left(1+\frac{h}{x}\right)}{\frac{h}{x}}\right\} \right\} }\)

Then, apply L'Hospital's Rule. (it's 0/0 now)

\(\color{#000000}{ \Large \displaystyle =\ln \left\{ \lim_{h\to 0}\exp\left\{\frac{ ~~~~\frac{- \frac{h}{x^2}}{1+\frac{h}{x}~}~~~~}{- \frac{h}{x^2}}\right\} \right\} }\)

\(\color{#000000}{ \displaystyle =\ln \left\{ \lim_{h\to 0}\exp\left\{1+\frac{h}{x}\right\} \right\} }\)

\(\color{#000000}{ \displaystyle =\ln \left\{ \exp\left\{1+\frac{0}{x}\right\} \right\} }\)

\(\color{#000000}{ \displaystyle =\ln \left\{ \exp\left\{1\right\} \right\}=\ln(e^1)=1 }\)

Thus, since,
\(\color{#000000}{ \displaystyle \lim_{h\to 0}\ln\left\{\left(\frac{x+h}{x}\right)^{x/h}\right\}=1}\)

Therefore,
\(\color{#000000}{ \displaystyle \frac{d}{dx}(\ln x)=\frac{1}{x} \lim_{h\to 0}\ln\left\{\left(\frac{x+h}{x}\right)^{x/h}\right\} =\frac{1}{x} \times 1 }\)

So,

\(\color{#000000}{ \displaystyle \frac{d}{dx}(\ln x)=\frac{1}{x} }\)

(Complete)

Answer 27

wow thank you

Answer 28

Yeah, but the route I usually take, if anyone wants to have everything proven, is to prove d/dx (e^x)=e^x, using the first principles.

\(\color{#000000}{ \displaystyle \frac{d}{dx}e^x=\lim_{h\to 0}\frac{e^{x+h}-e^x}{h}=\left(\lim_{h\to 0}e^x\right) \left(\lim_{h\to 0}\frac{e^h-1}{h} \right) =e^x\cdot 1 =e^x }\)

Where you will to show numerically (or using other approach than L'H'S) why
\(\color{#000000}{ \displaystyle \lim_{h\to 0}\frac{e^h-1}{h} =1 }\)

And then, I would post that prove for the derivative of ln(x), using the given derivative for e^x.

Answer 29

In any case, I think I am overtyping...
having been here in a while:)

Answer 30

If you need more material, let me know.

Answer 31

greatly appreciated. I was able to follow through the last line

so essentially,

\[y' = (\frac{ 1 }{ e^{y} })~and~e^{y} = x, so~y' = (\frac{ 1 }{ x })\]

we've just proven that since y = ln(x) that y' = (1/x)

Answer 32

Yup!