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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.

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