anonymous
  • anonymous
find dx/dy for the following equations 1.) y = square root of 3 (x+1)/x-1)
Mathematics
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
let me have a look to this closely
anonymous
  • anonymous
please tke your time
Owlfred
  • Owlfred
babz83 is smilbourn your mother?
amistre64
  • amistre64
cbrt((x+1)/(x-1)) right?
amistre64
  • amistre64
change the radical to its proper exponent form [ (x+1)/(x-1) ]^(1/3) and work thru
amistre64
  • amistre64
this aint gonna be nice with my formatting off
amistre64
  • amistre64
lets start with the ^1/3; power rule says bring it down and -1
anonymous
  • anonymous
@owlfred...nope remember she said she got an A in calc.... i would not be posting these questions
amistre64
  • amistre64
(1/3) [ (x+1)/(x-1) ]^(-2/3) and out pops the innards for the chain rule
amistre64
  • amistre64
Dx(x+1)/(x-1) = (x-1)(1) - (x+1)(-1) // (x-1)^2 by quotient rule
amistre64
  • amistre64
simplify to: 2x/(x-1)^2 right?
amistre64
  • amistre64
put it all together to go: 2x/3(x-1)^2) [ (x-1)/(x+1) ]^(2/3)
anonymous
  • anonymous
am following u amistre
anonymous
  • anonymous
let me just digest it!
amistre64
  • amistre64
\[\frac{2x}{3(x-1)^2}\left(\frac{x-1}{x+1}\right)^{2/3}\]right?
amistre64
  • amistre64
ill type it in the latex stuff fo try to make it easier to follow
anonymous
  • anonymous
nope
anonymous
  • anonymous
\[\sqrt[3]{x+1/x-1}\]
anonymous
  • anonymous
ah i get it thats after removing the cube root
amistre64
  • amistre64
\[\sqrt[3]{\frac{x+1}{x-1}}\] \[Dx\left(\frac{x+1}{x-1}\right)^{1/3}\] \[\frac{1}{3}\left(\frac{x+1}{x-1}\right)^{-2/3}Dx(\frac{x+1}{x-1})\] so far
amistre64
  • amistre64
-2/3 just flips it all over and goes positive again right?
anonymous
  • anonymous
right
amistre64
  • amistre64
\[\frac{1}{3}\left(\frac{x-1}{x+1}\right)^{2/3}Dx(\frac{x+1}{x-1})\] \[... Dx(\frac{x+1}{x-1})=\frac{bt'-b't}{b^2}\] \[... \frac{(x-1)(1)-(x+1)(-1)}{(x-1)^2}\] \[... \frac{(x-1)+(x+1)}{(x-1)^2}\] \[... \frac{x-1+x+1}{(x-1)^2}\] \[... \frac{2x}{(x-1)^2}\]
amistre64
  • amistre64
\[\frac{1}{3}\left(\frac{x-1}{x+1}\right)^{2/3}\frac{2x}{(x-1)^2}\] are all the parts, just simplify as wanted
amistre64
  • amistre64
i see a spot i missed; (x-1) has a derivative if 1, not -1
anonymous
  • anonymous
thnks a lot
anonymous
  • anonymous
i see i have a lot of serious reading to do
amistre64
  • amistre64
\[... \frac{(x-1)(1)-(x+1)(1)}{(x-1)^2}\] \[... \frac{x-1-x-1}{(x-1)^2}\] \[... \frac{-2}{(x-1)^2}\] \[\frac{1}{3}\left(\frac{x-1}{x+1}\right)^{2/3}\frac{-2}{(x-1)^2}\] thats proper, or should be better at least :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.