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kushashwa23
 3 years ago
Best ResponseYou've already chosen the best response.0let me have a look to this closely

Owlfred
 3 years ago
Best ResponseYou've already chosen the best response.0babz83 is smilbourn your mother?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1cbrt((x+1)/(x1)) right?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1change the radical to its proper exponent form [ (x+1)/(x1) ]^(1/3) and work thru

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1this aint gonna be nice with my formatting off

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1lets start with the ^1/3; power rule says bring it down and 1

babz83
 3 years ago
Best ResponseYou've already chosen the best response.0@owlfred...nope remember she said she got an A in calc.... i would not be posting these questions

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1(1/3) [ (x+1)/(x1) ]^(2/3) and out pops the innards for the chain rule

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1Dx(x+1)/(x1) = (x1)(1)  (x+1)(1) // (x1)^2 by quotient rule

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1simplify to: 2x/(x1)^2 right?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1put it all together to go: 2x/3(x1)^2) [ (x1)/(x+1) ]^(2/3)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1\[\frac{2x}{3(x1)^2}\left(\frac{x1}{x+1}\right)^{2/3}\]right?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1ill type it in the latex stuff fo try to make it easier to follow

babz83
 3 years ago
Best ResponseYou've already chosen the best response.0ah i get it thats after removing the cube root

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1\[\sqrt[3]{\frac{x+1}{x1}}\] \[Dx\left(\frac{x+1}{x1}\right)^{1/3}\] \[\frac{1}{3}\left(\frac{x+1}{x1}\right)^{2/3}Dx(\frac{x+1}{x1})\] so far

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.12/3 just flips it all over and goes positive again right?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1\[\frac{1}{3}\left(\frac{x1}{x+1}\right)^{2/3}Dx(\frac{x+1}{x1})\] \[... Dx(\frac{x+1}{x1})=\frac{bt'b't}{b^2}\] \[... \frac{(x1)(1)(x+1)(1)}{(x1)^2}\] \[... \frac{(x1)+(x+1)}{(x1)^2}\] \[... \frac{x1+x+1}{(x1)^2}\] \[... \frac{2x}{(x1)^2}\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1\[\frac{1}{3}\left(\frac{x1}{x+1}\right)^{2/3}\frac{2x}{(x1)^2}\] are all the parts, just simplify as wanted

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1i see a spot i missed; (x1) has a derivative if 1, not 1

babz83
 3 years ago
Best ResponseYou've already chosen the best response.0i see i have a lot of serious reading to do

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1\[... \frac{(x1)(1)(x+1)(1)}{(x1)^2}\] \[... \frac{x1x1}{(x1)^2}\] \[... \frac{2}{(x1)^2}\] \[\frac{1}{3}\left(\frac{x1}{x+1}\right)^{2/3}\frac{2}{(x1)^2}\] thats proper, or should be better at least :)
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