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Study23

I've been getting mixed answers from people. When do I know when I take the derivative of y to have yy' versus just y'???

  • one year ago
  • one year ago

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  1. Study23
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    Im referring to implicit differentiation...

    • one year ago
  2. amistre64
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    the derivative of y is y' either in implicit or explicit; the rules dont change because of the name of it

    • one year ago
  3. amistre64
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    spose you had y^2; then the derivaitvie is 2y y'

    • one year ago
  4. Study23
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    That's where I am confused. How do I know I should have yy', when in other cases you should have y'? In what cases do I only have y'?

    • one year ago
  5. amistre64
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    implicit is just the chain rule on steriods

    • one year ago
  6. amistre64
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    tell me, what is the derivative of say; x^3?

    • one year ago
  7. Study23
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    \(\ \Large 3x, \text{right?} \)

    • one year ago
  8. amistre64
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    no try it again, this time without the typo and ill tell you why your wrong again :)

    • one year ago
  9. Study23
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    \(\ \Large 3x^2 \text{ ?} \)

    • one year ago
  10. amistre64
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    thats better, but still wrong and heres why you are so used to taking a derivative "with respect to x" that you blindly assume that when the "with respect to" is not defined that it automatically has to be "with respect to x" since we dont know what this is in respect to, the chain rule pops out an x' until further notice

    • one year ago
  11. Study23
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    Hmmmmm......

    • one year ago
  12. Study23
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    Just x'? That's all??

    • one year ago
  13. amistre64
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    x^3 derives to 3x^2 x' now, when we define it as "with respect to x"; then x'=1

    • one year ago
  14. Study23
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    I thought you said we cant assume it to be the derivate with respect to x? Then we can't take the derivative of 3x^2?

    • one year ago
  15. amistre64
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    the point is, without knowing what it is derived with respect to .... the chain rule pops out a prime of the function what is the derivative of y? seeing that you know nothing of the rule that defines y

    • one year ago
  16. Study23
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    \(\ \large 1 \times y' \text{...? } \)

    • one year ago
  17. amistre64
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    correct

    • one year ago
  18. amistre64
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    what is the derivative of 3y^2 seeing that you know nothing about the rule that defines y

    • one year ago
  19. Study23
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    \(\ \text{Well, } 6y\times y' \text{ ? } \)

    • one year ago
  20. amistre64
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    think of all the variables as functions of some unknown variable 3 [y(?)]^2 and use the power and the chain rule

    • one year ago
  21. amistre64
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    3*2[y(?)]*y'(?) hence .... 6y y'

    • one year ago
  22. Study23
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    I see.

    • one year ago
  23. amistre64
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    when you start learning calc, they keep it simpla and only define things in terms of some known rules and it seems to put a mistaken idea in the heads about what is actually happening

    • one year ago
  24. Study23
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    Hmmm... Okay. Do you think you could help me with a implicit differentiation problem that I've been stuck on for like over an hour?

    • one year ago
  25. amistre64
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    sure

    • one year ago
  26. Study23
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    \(\ \Huge \text{THANKS! So, it is: } \)

    • one year ago
  27. Study23
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    \(\ \Large \frac{x^2}{16}-\frac{y^2}{9} = 1. \text{Find the tangent line to the curve, }\) \(\ \Large \text{ To the point} (-5, \frac{9}{4}. ) \)

    • one year ago
  28. amistre64
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    as long as my internet connection doesnt conk out on me

    • one year ago
  29. amistre64
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    power rule, constant rule, and the obiquitous chain rule ...

    • one year ago
  30. Study23
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    So like 2x/16? Doesn't the 16 become 0?

    • one year ago
  31. amistre64
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    \[\frac{1}{16}x^2=\frac{x^2}{16}\]

    • one year ago
  32. amistre64
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    the 16 doesnt become 0

    • one year ago
  33. Study23
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    I see now

    • one year ago
  34. Study23
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    So for the y, Ill take a shot at it: \(\ \Large -\frac{2yy'}{9} ? \)

    • one year ago
  35. amistre64
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    go ahead and pop in an x' just to keep things kosher, in the end x'=1 but its good to root that in your mindset

    • one year ago
  36. amistre64
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    and so far, so good

    • one year ago
  37. Study23
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    I don't think Ive ever heard that one before "to keep things kosher..." ;)

    • one year ago
  38. Study23
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    So now I simplify? \(\ \Large \frac{2xx'}{16} \rightarrow \frac{x}{8} ? \)

    • one year ago
  39. Study23
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    As for the y, Im stuck now.

    • one year ago
  40. amistre64
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    well, we know that 1'=0 for the constant, so lets write up the derivative that we have so far

    • one year ago
  41. Study23
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    Btw, what allows us to say x'=1?

    • one year ago
  42. amistre64
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    im sure that somewhere in the instructions its asking you to find dy/dx

    • one year ago
  43. amistre64
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    the phrase implicit even suggests that y is a function of something ... and at this stage they infer that that something is rule invloving x

    • one year ago
  44. Study23
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    Yea, notspecifically in this set of problems, but in a previous set it did

    • one year ago
  45. amistre64
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    so, write up the derivate that we have so far, then the rest should become clear

    • one year ago
  46. Study23
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    \(\ \Large \frac{x}{8} - \frac{2yy'}{9} = 0 \)

    • one year ago
  47. amistre64
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    and the given point tells us the values for x and y, so plug those in as well

    • one year ago
  48. Study23
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    What about the y'? Do I need to isolate it??

    • one year ago
  49. amistre64
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    yes, but i was just pointing out that y' was going to be the only unknown to solve for :)

    • one year ago
  50. amistre64
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    and that y' IS the slope of thhe tangent line at the given point

    • one year ago
  51. Study23
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    Okay. :) SO I substituted the point in, and I get y'=\(\ -\frac{41}{72} \)

    • one year ago
  52. amistre64
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    ill trust you on the math for that one; now how do we define a line if we know th eslope and a point on the line?

    • one year ago
  53. Study23
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    Slope-Point Formula! O R Or as it is more commonly known as, Point-Slope Form

    • one year ago
  54. amistre64
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    correct, and dbl chk that y' -5, 9/4 -5/8 - y'/4 = 0 --> y' = -4(5/8) right?

    • one year ago
  55. Study23
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    Isnt it -9/4? Not y'/4? What happened to that y?

    • one year ago
  56. amistre64
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    2/9 * 9/4 = 1/2 so, -2(5/8) = -10/8

    • one year ago
  57. Study23
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    9/4

    • one year ago
  58. amistre64
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    \[y=\frac94\]\[\frac29yy'=\frac29\frac94y'=\frac12y'\]

    • one year ago
  59. Study23
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    Okay, that makes sense. That slope is correct btw per the book!

    • one year ago
  60. amistre64
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    :)

    • one year ago
  61. Study23
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    Okay, @amistre64 It looks good now! Thank you for your help the past hour! \(\ \Large \text{Words cannot express my utmost appreciation for your help!} \) \(\ \Large \text{Thank You, THANK you, THANK YOU SO MUCH!!!!! } \)

    • one year ago
  62. amistre64
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    my internet connection is toast, so ill have to say gnite :) good luck

    • one year ago
  63. Study23
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    :D

    • one year ago
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