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.

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