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MTALHAHASSAN2

  • one year ago

Differentiate each of the following function: 2x^3-1/x^2

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  1. MTALHAHASSAN2
    • one year ago
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    F1(x)= 3(2x^2-1)(x^2)+(2x^3-1)(2x)

  2. MTALHAHASSAN2
    • one year ago
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    So the derivative of the given function be like that right

  3. campbell_st
    • one year ago
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    well I'd start by writing both terms using index notation \[y = 2x^3 - x^{-2}\] now apply the rule for differentiation to both terms \[if~~ y = x^n ~~~y' = nx^{n - 1}\]

  4. MTALHAHASSAN2
    • one year ago
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    Oh ok but you miss the -1

  5. MTALHAHASSAN2
    • one year ago
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    2x^3-1

  6. campbell_st
    • one year ago
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    no I didn't I used index notation \[\frac{1}{x^2} = x^{-2}\]

  7. campbell_st
    • one year ago
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    so is the question 1. \[y = 2x^3 - \frac{1}{x^2}\] or 2. \[y = 2x^{3 - \frac{1}{x^2}}\]

  8. MTALHAHASSAN2
    • one year ago
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    Both are wrong

  9. MTALHAHASSAN2
    • one year ago
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    X^2 also a denominater of 2x^3

  10. MTALHAHASSAN2
    • one year ago
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    @campbell_st

  11. campbell_st
    • one year ago
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    ok... so its \[\frac{2x^3 -1}{x^2}\]

  12. MTALHAHASSAN2
    • one year ago
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    Yes

  13. MTALHAHASSAN2
    • one year ago
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    F1(x)= 3(2x^2-1)(x^2)+(2x^3-1)(2x)

  14. campbell_st
    • one year ago
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    ok so there are a couple of ways to do this... 1. use the quitient rule... or 2. split it into 2 fractions... \[y = \frac{2x^3}{x^2} - \frac{1}{x^2}\] so simplifying and using index notation you get \[y = 2x - x^{-2}\] so the derivative is \[y' = 2 + \frac{2}{x^3}\]

  15. campbell_st
    • one year ago
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    using the method you ahve attempted, which looks like the quotient rule you forgot the denominator if \[y = \frac{f(x)}{g(x)} \] then \[y' = \frac{g(x) f'(x) - f(x) g'(x)}{(g(x))^2}\]

  16. MTALHAHASSAN2
    • one year ago
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    No I guess that is product rule

  17. MTALHAHASSAN2
    • one year ago
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    Y′=f(x)1h(x)+f(x)h(x)1

  18. MTALHAHASSAN2
    • one year ago
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    @campbell_st

  19. campbell_st
    • one year ago
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    well if you are using the product rule you need to rewrite it using negative powers for the denominator \[\frac{2x^3 -1}{x^2} = (2x^3 -1)(x^{-2})\] which will make a big difference

  20. MTALHAHASSAN2
    • one year ago
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    Oh ok but can I use any of them

  21. MTALHAHASSAN2
    • one year ago
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    I mean the rules

  22. campbell_st
    • one year ago
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    you can use any of the rules, product rule, but careful with the power in the denominator quotient rule... normal diffierentiation rule after simplifying... they will all give the same answer

  23. MTALHAHASSAN2
    • one year ago
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    Wait then what is ur answer

  24. MTALHAHASSAN2
    • one year ago
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    F1(x)= 3(2x^2-1)(x)^-2+(2x^3-1)-2x

  25. MTALHAHASSAN2
    • one year ago
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    F1(x)=6x^-2(x^2-1)-4x(x^3-1)

  26. MTALHAHASSAN2
    • one year ago
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    @campbell_st

  27. anonymous
    • one year ago
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    You're making it harder than it is

  28. anonymous
    • one year ago
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    \[y=\frac{2x^3-1}{x^2} \iff y = (2x^3-1)(x^{-2})\]

  29. anonymous
    • one year ago
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    \[y'=2\cdot 3x^2 \cdot x^{-2} + (-2x^{-3})(2x^3-1)\]\[y'=2\cdot 3x^{2-2} -2x^{-3}(2x^3)+2x^{-3}\]\[y'=6x^{0}-(2\cdot 2)x^{0} +2x^{-3}\]\[y'=6-4+2x^{-3}\]\[y'= ...

  30. anonymous
    • one year ago
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    \[y' =....\] **

  31. triciaal
    • one year ago
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    do you have any questions?

  32. MTALHAHASSAN2
    • one year ago
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    Can someone plz explain this question again to me??

  33. triciaal
    • one year ago
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    suggest you use parenthesis to make sure your question is clear

  34. triciaal
    • one year ago
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    |dw:1444369433856:dw|

  35. MTALHAHASSAN2
    • one year ago
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    I am having trouble after that positive sign.

  36. triciaal
    • one year ago
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    3 approaches with practice you can decide which is better for you. end with the same answer.

  37. MTALHAHASSAN2
    • one year ago
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    I will go with easier one

  38. triciaal
    • one year ago
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    1) the quotient rule 2)rewrite with index use product rule and chain rule 3) divide and do each term

  39. MTALHAHASSAN2
    • one year ago
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    Wait what you mean by the index

  40. triciaal
    • one year ago
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    that's the point they are all easy. with practice you will determine when to use which one or you can just stick to one format. personal preference.

  41. anonymous
    • one year ago
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    I used the product rule: \(f' \cdot g + g' \cdot f\)

  42. MTALHAHASSAN2
    • one year ago
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    Can you please explain the second one

  43. triciaal
    • one year ago
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    what zepdrix did and also campbell above it's rewriting to use the product rule

  44. MTALHAHASSAN2
    • one year ago
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    I know the product one

  45. triciaal
    • one year ago
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    |dw:1444370121922:dw|

  46. MTALHAHASSAN2
    • one year ago
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    Got u

  47. triciaal
    • one year ago
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    good

  48. MTALHAHASSAN2
    • one year ago
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    So lets us the product on e

  49. triciaal
    • one year ago
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    |dw:1444370231542:dw|

  50. MTALHAHASSAN2
    • one year ago
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    Ok, looks good

  51. triciaal
    • one year ago
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    |dw:1444370312351:dw|

  52. MTALHAHASSAN2
    • one year ago
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    Ok

  53. triciaal
    • one year ago
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    so you just need to know how to apply the rules and practice

  54. triciaal
    • one year ago
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    additional note you can use substitution to apply the chain rule

  55. triciaal
    • one year ago
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    any final questions on this?

  56. triciaal
    • one year ago
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    ok goodnight

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