Mathematics
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

if not we can derive that too!

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

is it can derivate with limit equation???

OpenStudy (anonymous):

no usually done this way. chain rule problem.

OpenStudy (anonymous):

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 $\frac{d}{dx}\ln(x)=\frac{1}{x}$

OpenStudy (anonymous):

this works for finding the derivative of any inverse function

OpenStudy (anonymous):

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 $\frac{d}{dx}f'(x)=\frac{1}{f'(f^{-1}(x))}$

OpenStudy (anonymous):

is it only definition?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

why ln (x) is inverse of $e^{x}$

OpenStudy (anonymous):

oh i'm understand thanks satelite

OpenStudy (anonymous):

but can derivate ln (x) with limit equation?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Here's a version using limits: http://www.math.com/tables/derivatives/more/ln.htm

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

now you are supposed to take the limit inside? hmmm

OpenStudy (anonymous):

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.

Latest Questions
Victorya: why do you use math in this world?
54 minutes ago 10 Replies 5 Medals
Gorillaz1002: i need your opinion on this i updated it a little
2 hours ago 23 Replies 3 Medals
MrsHero: Help ss down below ;/..
2 hours ago 7 Replies 1 Medal
MrsHero: Help ss down below ;/
2 hours ago 0 Replies 0 Medals