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babz83

  • 3 years ago

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

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  1. babz83
    • 3 years ago
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    \[\sqrt[3]{X+1/x-1}\]

  2. babz83
    • 3 years ago
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    2) y = cos(lnX)

  3. kushashwa23
    • 3 years ago
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    ok

  4. kushashwa23
    • 3 years ago
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    let me have a look to this closely

  5. babz83
    • 3 years ago
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    please tke your time

  6. Owlfred
    • 3 years ago
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    babz83 is smilbourn your mother?

  7. amistre64
    • 3 years ago
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    cbrt((x+1)/(x-1)) right?

  8. amistre64
    • 3 years ago
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    change the radical to its proper exponent form [ (x+1)/(x-1) ]^(1/3) and work thru

  9. amistre64
    • 3 years ago
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    this aint gonna be nice with my formatting off

  10. amistre64
    • 3 years ago
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    lets start with the ^1/3; power rule says bring it down and -1

  11. babz83
    • 3 years ago
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    @owlfred...nope remember she said she got an A in calc.... i would not be posting these questions

  12. amistre64
    • 3 years ago
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    (1/3) [ (x+1)/(x-1) ]^(-2/3) and out pops the innards for the chain rule

  13. amistre64
    • 3 years ago
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    Dx(x+1)/(x-1) = (x-1)(1) - (x+1)(-1) // (x-1)^2 by quotient rule

  14. amistre64
    • 3 years ago
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    simplify to: 2x/(x-1)^2 right?

  15. amistre64
    • 3 years ago
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    put it all together to go: 2x/3(x-1)^2) [ (x-1)/(x+1) ]^(2/3)

  16. babz83
    • 3 years ago
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    am following u amistre

  17. babz83
    • 3 years ago
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    let me just digest it!

  18. amistre64
    • 3 years ago
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    \[\frac{2x}{3(x-1)^2}\left(\frac{x-1}{x+1}\right)^{2/3}\]right?

  19. amistre64
    • 3 years ago
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    ill type it in the latex stuff fo try to make it easier to follow

  20. babz83
    • 3 years ago
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    nope

  21. babz83
    • 3 years ago
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    \[\sqrt[3]{x+1/x-1}\]

  22. babz83
    • 3 years ago
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    ah i get it thats after removing the cube root

  23. amistre64
    • 3 years ago
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    \[\sqrt[3]{\frac{x+1}{x-1}}\] \[Dx\left(\frac{x+1}{x-1}\right)^{1/3}\] \[\frac{1}{3}\left(\frac{x+1}{x-1}\right)^{-2/3}Dx(\frac{x+1}{x-1})\] so far

  24. amistre64
    • 3 years ago
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    -2/3 just flips it all over and goes positive again right?

  25. babz83
    • 3 years ago
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    right

  26. amistre64
    • 3 years ago
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    \[\frac{1}{3}\left(\frac{x-1}{x+1}\right)^{2/3}Dx(\frac{x+1}{x-1})\] \[... Dx(\frac{x+1}{x-1})=\frac{bt'-b't}{b^2}\] \[... \frac{(x-1)(1)-(x+1)(-1)}{(x-1)^2}\] \[... \frac{(x-1)+(x+1)}{(x-1)^2}\] \[... \frac{x-1+x+1}{(x-1)^2}\] \[... \frac{2x}{(x-1)^2}\]

  27. amistre64
    • 3 years ago
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    \[\frac{1}{3}\left(\frac{x-1}{x+1}\right)^{2/3}\frac{2x}{(x-1)^2}\] are all the parts, just simplify as wanted

  28. amistre64
    • 3 years ago
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    i see a spot i missed; (x-1) has a derivative if 1, not -1

  29. babz83
    • 3 years ago
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    thnks a lot

  30. babz83
    • 3 years ago
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    i see i have a lot of serious reading to do

  31. amistre64
    • 3 years ago
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    \[... \frac{(x-1)(1)-(x+1)(1)}{(x-1)^2}\] \[... \frac{x-1-x-1}{(x-1)^2}\] \[... \frac{-2}{(x-1)^2}\] \[\frac{1}{3}\left(\frac{x-1}{x+1}\right)^{2/3}\frac{-2}{(x-1)^2}\] thats proper, or should be better at least :)

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