Question

lim h->0
(ln(x+h)-ln(x))/h

Answer 1

what is the f(x) ?

Answer 2

you are trying to find the derivative of ln(x) using the first principles ?

Answer 3

\(\Large\color{black}{\displaystyle\lim_{h \rightarrow ~0}\frac{\ln(x+h)-\ln(x)}{h}}\)

Answer 4

yes

Answer 5

are you required to use this rule specifically, or have you learned the basic derivative such as derivative of e^x?

Answer 6

I've learned "a special limit"

Answer 7

okay, sorry for late reply. I would just maybe do it as it is, if you have to use limits.
first apply, the
ln(a)-ln(b) = ln(a/b).

Answer 8

I did learn e^x actually, I'm just really lost on everything right now and I forgot what that was, but I'm supposed to be past that part of the book. Sorry. I just really don't understand any of this.

Answer 9

So, we maybe can take the derivative of ln(x) as.

y=ln(x)

e^y=x

differentiate both sides with respect to x,

(dy/dx) e^y = 1

then divide both sides by e^y, and substitute x for e^y, since you know that e^y=x from the beginning.

Answer 10

for some reason this seems easier, than doing a limit change twice.

Answer 11

can you show me that...I'm really, really lost on everything.

Answer 12

which way you want to go, do the limit?

Answer 13

\(\Large\color{black}{\displaystyle\lim_{h \rightarrow ~0}\frac{\ln(x+h)-\ln(x)}{h}}\)


\(\Large\color{black}{\displaystyle\lim_{h \rightarrow ~0}\frac{\ln(\frac{x+h}{x})}{h}}\)

does this step look familiar?

Answer 14

I don't know...shouldn't the ln((x+h)/2) become 1/abs((x+h)/2)

Answer 15

why is it over 2?

Answer 16

I meant x

Answer 17

Idk, I actually haven't heard of such a rule.

Answer 18

ok, probably not then

Answer 19

I might not now all the rules, but it might be the correct rule:
the way I was going to do it is:

\(\Large\color{black}{\displaystyle\lim_{h \rightarrow ~0}\frac{\ln(\frac{x+h}{x})}{h}}\)

\(\Large\color{black}{\displaystyle\lim_{h \rightarrow ~0}\frac{\ln(1+\frac{h}{x})}{h}}\)

then, let u=h/x.... and on.

Answer 20

not *know...

Answer 21

where did you get u?

Answer 22

I am doing substitution to change the limit.

Answer 23

ok

Answer 24

I am a very slow helper apparently, maybe you can tag someone who can better help you with this?

Answer 25

I mean it is up to you.

Answer 26

no, I'm just really slow at understanding. It's not you.

Answer 27

Well, it is a hard topic....

\(\Large\color{black}{\displaystyle\lim_{h \rightarrow ~0}\frac{\ln(1+\frac{h}{x})}{h}}\) can you tell me what do we get
when we substitute u for h/x ?

Answer 28

(ln(1+u))/h
but what do you do with h?

Answer 29

the h becomes ux.
and h->0 gives us that u->0.

Answer 30

\(\Large\color{black}{\displaystyle\lim_{h \rightarrow ~0}\frac{\ln(1+\frac{h}{x})}{h}}\)

changing the limit, we get:

\(\Large\color{black}{\displaystyle\lim_{u \rightarrow ~0}\frac{\ln(1+u)}{ux}}\)

Answer 31

then you can take 1/x outside the limit.

Answer 32

I don't know what you're saying

Answer 33

you don't know what I mean by take the 1/x out, or don't know how I changed the limit?

Answer 34

both

Answer 35

if you don't know how to change the limit, we can take the derivative of ln(x) the way I initially proposed.

\(\large\color{black}{y= \ln(x) }\)

\(\large\color{black}{y= \log_e(x) }\)

\(\large\color{black}{e^y=x }\)

and on as I did before....

Answer 36

ok, yeah

Answer 37

So, you know that the derivative of e^x is just e^x (if you want we can prove that too)

Answer 38

yeah I know that

Answer 39

okay, but the derivative (with respect to x) of e^y will have the chain rule for (dy/dx)

(dy/dx is basically same as y', so which ever notation you like better that will be what we are going to use)

So differentiating e^y we get (dy/dx) * e^y, and differentiating x we get just 1.

So the derivative of e^y=x is:

(dy/dx) e^y = 1

then we need to solve for the dy/dx, 9dividing on e^y on both sides,

(dy/dx)= 1 / e^y

then we know (from the beginning) that e^y=x, so substitute x instead of "e^y".

you get that;

(dy/dx) = ?

Answer 40

I'll post a prove for the derivative of e^x in just a second: while you read what i wrote.

Answer 41

\(\Large\color{black}{\displaystyle\lim_{h \rightarrow ~0}\frac{e^{x+h}-e^x}{h}}\)


\(\Large\color{black}{\displaystyle\lim_{h \rightarrow ~0}\frac{e^{x}e^h-e^x}{h}}\)


\(\Large\color{black}{\displaystyle\lim_{h \rightarrow ~0}\frac{e^{x}(1-e^h)}{h}}\)


\(\Large\color{black}{\displaystyle\lim_{h \rightarrow ~0}\frac{1-e^h}{h} \times \displaystyle\lim_{h \rightarrow ~0}e^{x}}\)


\(\Large\color{black}{\displaystyle\lim_{h \rightarrow ~0}\frac{1-e^h}{h} \times e^{x}}\)

we know that: \(\Large\color{black}{\displaystyle\lim_{h \rightarrow ~0}\frac{1-e^h}{h}=1}\) (we can prove this too if you want to)

\(\Large\color{black}{1 \times e^{x}}\)

\(\Large\color{black}{e^{x}}\)

Answer 42

ok

Answer 43

(dy/dx)= 1/x

Answer 44

yes that is right, so the derivative of ln(x) is 1/x.

Answer 45

got it

Answer 46

thank you

Answer 47

you welcome!