Question

can you help me to show d(ln(x))/dx=1/x.

Answer 1

sure, so long as you get to use
\[\frac{d}{dx}f^{-1}(x)=\frac{1}{f'(f(x))}\]

Answer 2

if not we can derive that too!

Answer 3

we can start by noting that
\[e^{\ln(x)}=x\]

Answer 4

then take the derivative of both sides. the left hand side is easy, the derivative of
\[x\] is 1

Answer 5

the derivative of
\[e^{\ln(x)}=e^{\ln(x)}\times \frac{d}{dx}\ln(x)\] by the chain rule, so we know that
\[e^{\ln(x)}\times \frac{d}{dx}\ln(x)=1\]

Answer 6

is it can derivate with limit equation???

Answer 7

no usually done this way. chain rule problem.

Answer 8

this line
\[e^{\ln(x)}\times \frac{d}{dx}\ln(x)=1\]is the same as saying
\[x\times \frac{d}{dx}\ln(x)=1\] so solving we get
\[\frac{d}{dx}\ln(x)=\frac{1}{x}\]

Answer 9

this works for finding the derivative of any inverse function

Answer 10

you repeat the process with
\[f(f^{-1}(x))=x\] take the derivative via the chain rule and get
\[f'(f^{-1}(x))\frac{d}{dx}f'(x)=1\] and so
\[\frac{d}{dx}f'(x)=\frac{1}{f'(f^{-1}(x))}\]

Answer 11

is it only definition?

Answer 12

we are using the chain rule. it is not by definition. at this point your definition of
\[\ln(x)\] is probably the inverse of
\[e^x\]

Answer 13

so the method outlined above shows how to take the derivative of any inverse function

Answer 14

you can use if for
\[\sin^{-1}(x)\] as well

Answer 15

if your definition of the natural log is
\[\int_1^x \frac{1}{t}dt\] the the derivative is automatically
\[\frac{1}{x}\] but my guess is that you don't use that definition for the log

Answer 16

why ln (x) is inverse of \[e^{x}\]

Answer 17

oh i'm understand thanks satelite

Answer 18

but can derivate ln (x) with limit equation?

Answer 19

it really depends on your definition of
\[\ln(x)\]
how to compute
\[\lim_{h\rightarrow 0}\frac{\ln(x+h)-\ln(x)}{h}\]

Answer 20

Here's a version using limits:

http://www.math.com/tables/derivatives/more/ln.htm

Answer 21

if you define
\[\ln(x)=\int_1^x\frac{dt}{t}\] then as i said it is immediate since the derivative of the integral is the integrand.

Answer 22

let me see if i get this.
\[\lim_{h\rightarrow 0}\frac{\ln(x+h)-\ln(x)}{h}\]
\[=\lim_{h\rightarrow 0}\frac{\ln(\frac{x+h}{x})}{h}\]
\[=\lim_{h\rightarrow 0}\frac{\ln(1+\frac{h}{x})}{h}\]

Answer 23

\[=\lim_{h\rightarrow 0}\frac{1}{h}\ln(1+\frac{x}{h})\]
\[=\lim_{h \rightarrow 0}\ln(1+\frac{x}{h})^{\frac{1}{h}}\]

Answer 24

now you are supposed to take the limit inside? hmmm

Answer 25

looks like some slight of hand to me but i guess so. easier to use the chain rule and method for taking derivative of an inverse
oh and a change of variable too.