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anonymous

  • 5 years ago

can some one help me diferentiate x^1/2+x^-4/3

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  1. anonymous
    • 5 years ago
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    in respect to what?

  2. anonymous
    • 5 years ago
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    to y

  3. anonymous
    • 5 years ago
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    \[y = x ^{1/2} + x ^{-4/3} \] is this the original function?

  4. anonymous
    • 5 years ago
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    no the original function is y=sqrt of x + 1/cube root of x^4

  5. anonymous
    • 5 years ago
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    write it as an equation because now im conufse

  6. anonymous
    • 5 years ago
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    \[\sqrt{x}+1/\sqrt[3]{x^4}\]

  7. anonymous
    • 5 years ago
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    dy/dx = 1/2 x^(-1/2) -4/3 x^-7/3

  8. anonymous
    • 5 years ago
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    rewrite it as \[y= x ^{1/2} + 1 /(x ^{4/3})\] \[y= x ^{1/2} + (x ^{-4/3})\] \[dy/dx= (1/2)x ^{-1/2} + (-4/3)x ^{-7/3}\]

  9. anonymous
    • 5 years ago
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    understand how to get it?

  10. anonymous
    • 5 years ago
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    thats the first answer i got but i dont uderstand how i got it

  11. anonymous
    • 5 years ago
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    i understand the first step but am lost at the second

  12. anonymous
    • 5 years ago
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    are we using both the product rule and quotient rule here?

  13. anonymous
    • 5 years ago
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    no, just differentiate each term separately

  14. anonymous
    • 5 years ago
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    its the chain rule for example when you find the derivative of \[x^a\] the answer is \[a * x ^{a-1}\]

  15. anonymous
    • 5 years ago
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    @Romero, I dont think its the chain rule since this is not a composite function this question only requires you to do basic differentiation using the rule "y=(a)x^n" differentiates to "dy/dx= (a)(n)x^(n-1)

  16. anonymous
    • 5 years ago
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    sorry for not getting the word right but the concept is the same in finding the derivative.

  17. anonymous
    • 5 years ago
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    see i actually got what romero got but tried to the product rule after

  18. anonymous
    • 5 years ago
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    You dont need to use the product or quotient rule here. That's only when you have 2 variables multiplying to dividing.

  19. anonymous
    • 5 years ago
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    yeah i just saw that

  20. anonymous
    • 5 years ago
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    thank you so much man i really appreciate it trying to cram for exam am going to have in a couple of days and this really helps

  21. anonymous
    • 5 years ago
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    Sure thing man. I think the problem you're having is rewriting out square roots. Here this might help.\[\sqrt[a]{x^b} = x ^{(b/a)}\] if you have no number for a then a =2 also \[1/x^a = 1*x ^{-a}\] It was hard for me to see it at first but in time this will become second hand to you.

  22. anonymous
    • 5 years ago
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    do you think with couple of more problems i think i got how to do them but i dont know if they are right

  23. anonymous
    • 5 years ago
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    sure

  24. anonymous
    • 5 years ago
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    y=cos(tanx) have to deferentiate

  25. anonymous
    • 5 years ago
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    you use the chain rule so you should get y'= sin(tan x) * sec^2 x

  26. anonymous
    • 5 years ago
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    why do i use the product rule here

  27. anonymous
    • 5 years ago
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    no you dont use the product rule you only have one x within tan that happens to be inside cos

  28. anonymous
    • 5 years ago
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    actually its suppose to be y'= -sin(tan x) * sec^2 x

  29. anonymous
    • 5 years ago
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    With this problem you need to use the chain rule because its a composite function. The chain rule is, if y=f(u) \[dy/dx = dy/du*du/dx\] Let u=tanx y=cos(u) differentiates to -sin(u) tanx differentiates to \[\sec ^{2}x\] therefore dy/dx= -sin(tanx)*\[\sec ^{2}x\]

  30. anonymous
    • 5 years ago
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    when y= cos(a) y' = cos(a)' * a' in this case a = tan(x) so y'= - sin(tan(x)) * tan(x)' where tan(x)'= sec^2 (x) so y'= - sin (tanx) * sec^2 x

  31. anonymous
    • 5 years ago
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    oh ok i am getting it

  32. anonymous
    • 5 years ago
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    how about 3x-2/\[\sqrt{2x+1}\]

  33. anonymous
    • 5 years ago
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    Use the quotient rule.

  34. anonymous
    • 5 years ago
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    how do i diferentiate \[\sqrt{2x+1}\]

  35. anonymous
    • 5 years ago
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    y= a/b where y'= (a' * b - b' * a)/ b^2 in this case a = 3x -2 \[b= \sqrt{2x + 1}\]

  36. anonymous
    • 5 years ago
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    so a'=3 right

  37. anonymous
    • 5 years ago
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    yes a'=3, b'=1/2(2x+1)^(-1/2)

  38. anonymous
    • 5 years ago
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    so am getting 3/2(2x+1)^-1/2-(2x+1)^-1/2 over 2x+1 is that right

  39. anonymous
    • 5 years ago
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    I'm not too sure if that's right maybe im just confused with the typing but I got 3(2x+1)^(1/2) - (3x+2)(2x+1)^(-1/2) all over 2x+1

  40. anonymous
    • 5 years ago
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    I found this really good summary of differentiation formulaes, it covers up most basic differentiation topics. Hope you find it useful for your exam coming up soon. :)

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  41. anonymous
    • 5 years ago
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    AWESOME THANKS MAN

  42. anonymous
    • 5 years ago
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    how did you get rid of the 1/2 after you derived \[\sqrt{2x+1}\]

  43. anonymous
    • 5 years ago
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    when you differentiate root 2x+1, you use the chain rule. so its \[\sqrt{2x+1}, dy/dx= (1/2)(2)(2x+1)^{-1/2}\] so the 2 cancels out with the 1/2

  44. anonymous
    • 5 years ago
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    ok cool

  45. anonymous
    • 5 years ago
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    thnxs for the help man

  46. anonymous
    • 5 years ago
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    no problemo ;) btw, good luck for yur exam

  47. anonymous
    • 5 years ago
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    thns man am gonna need it

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