find dx/dy for the following equations 1.) y = square root of 3 (x+1)/x-1)

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find dx/dy for the following equations 1.) y = square root of 3 (x+1)/x-1)

Mathematics
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\[\sqrt[3]{X+1/x-1}\]
2) y = cos(lnX)
ok

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let me have a look to this closely
please tke your time
babz83 is smilbourn your mother?
cbrt((x+1)/(x-1)) right?
change the radical to its proper exponent form [ (x+1)/(x-1) ]^(1/3) and work thru
this aint gonna be nice with my formatting off
lets start with the ^1/3; power rule says bring it down and -1
@owlfred...nope remember she said she got an A in calc.... i would not be posting these questions
(1/3) [ (x+1)/(x-1) ]^(-2/3) and out pops the innards for the chain rule
Dx(x+1)/(x-1) = (x-1)(1) - (x+1)(-1) // (x-1)^2 by quotient rule
simplify to: 2x/(x-1)^2 right?
put it all together to go: 2x/3(x-1)^2) [ (x-1)/(x+1) ]^(2/3)
am following u amistre
let me just digest it!
\[\frac{2x}{3(x-1)^2}\left(\frac{x-1}{x+1}\right)^{2/3}\]right?
ill type it in the latex stuff fo try to make it easier to follow
nope
\[\sqrt[3]{x+1/x-1}\]
ah i get it thats after removing the cube root
\[\sqrt[3]{\frac{x+1}{x-1}}\] \[Dx\left(\frac{x+1}{x-1}\right)^{1/3}\] \[\frac{1}{3}\left(\frac{x+1}{x-1}\right)^{-2/3}Dx(\frac{x+1}{x-1})\] so far
-2/3 just flips it all over and goes positive again right?
right
\[\frac{1}{3}\left(\frac{x-1}{x+1}\right)^{2/3}Dx(\frac{x+1}{x-1})\] \[... Dx(\frac{x+1}{x-1})=\frac{bt'-b't}{b^2}\] \[... \frac{(x-1)(1)-(x+1)(-1)}{(x-1)^2}\] \[... \frac{(x-1)+(x+1)}{(x-1)^2}\] \[... \frac{x-1+x+1}{(x-1)^2}\] \[... \frac{2x}{(x-1)^2}\]
\[\frac{1}{3}\left(\frac{x-1}{x+1}\right)^{2/3}\frac{2x}{(x-1)^2}\] are all the parts, just simplify as wanted
i see a spot i missed; (x-1) has a derivative if 1, not -1
thnks a lot
i see i have a lot of serious reading to do
\[... \frac{(x-1)(1)-(x+1)(1)}{(x-1)^2}\] \[... \frac{x-1-x-1}{(x-1)^2}\] \[... \frac{-2}{(x-1)^2}\] \[\frac{1}{3}\left(\frac{x-1}{x+1}\right)^{2/3}\frac{-2}{(x-1)^2}\] thats proper, or should be better at least :)

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