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Study23 Group Title

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'???

  • 2 years ago
  • 2 years ago

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  1. Study23 Group Title
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    Im referring to implicit differentiation...

    • 2 years ago
  2. amistre64 Group Title
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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

    • 2 years ago
  3. amistre64 Group Title
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    spose you had y^2; then the derivaitvie is 2y y'

    • 2 years ago
  4. Study23 Group Title
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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'?

    • 2 years ago
  5. amistre64 Group Title
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    implicit is just the chain rule on steriods

    • 2 years ago
  6. amistre64 Group Title
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    tell me, what is the derivative of say; x^3?

    • 2 years ago
  7. Study23 Group Title
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    \(\ \Large 3x, \text{right?} \)

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

    • 2 years ago
  9. Study23 Group Title
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    \(\ \Large 3x^2 \text{ ?} \)

    • 2 years ago
  10. amistre64 Group Title
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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

    • 2 years ago
  11. Study23 Group Title
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    Hmmmmm......

    • 2 years ago
  12. Study23 Group Title
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    Just x'? That's all??

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

    • 2 years ago
  14. Study23 Group Title
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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?

    • 2 years ago
  15. amistre64 Group Title
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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

    • 2 years ago
  16. Study23 Group Title
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    \(\ \large 1 \times y' \text{...? } \)

    • 2 years ago
  17. amistre64 Group Title
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    correct

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

    • 2 years ago
  19. Study23 Group Title
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    \(\ \text{Well, } 6y\times y' \text{ ? } \)

    • 2 years ago
  20. amistre64 Group Title
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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

    • 2 years ago
  21. amistre64 Group Title
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    3*2[y(?)]*y'(?) hence .... 6y y'

    • 2 years ago
  22. Study23 Group Title
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    I see.

    • 2 years ago
  23. amistre64 Group Title
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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

    • 2 years ago
  24. Study23 Group Title
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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?

    • 2 years ago
  25. amistre64 Group Title
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    sure

    • 2 years ago
  26. Study23 Group Title
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    \(\ \Huge \text{THANKS! So, it is: } \)

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

    • 2 years ago
  28. amistre64 Group Title
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    as long as my internet connection doesnt conk out on me

    • 2 years ago
  29. amistre64 Group Title
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    power rule, constant rule, and the obiquitous chain rule ...

    • 2 years ago
  30. Study23 Group Title
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    So like 2x/16? Doesn't the 16 become 0?

    • 2 years ago
  31. amistre64 Group Title
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    \[\frac{1}{16}x^2=\frac{x^2}{16}\]

    • 2 years ago
  32. amistre64 Group Title
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    the 16 doesnt become 0

    • 2 years ago
  33. Study23 Group Title
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    I see now

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

    • 2 years ago
  35. amistre64 Group Title
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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

    • 2 years ago
  36. amistre64 Group Title
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    and so far, so good

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

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

    • 2 years ago
  39. Study23 Group Title
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    As for the y, Im stuck now.

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

    • 2 years ago
  41. Study23 Group Title
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    Btw, what allows us to say x'=1?

    • 2 years ago
  42. amistre64 Group Title
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    im sure that somewhere in the instructions its asking you to find dy/dx

    • 2 years ago
  43. amistre64 Group Title
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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

    • 2 years ago
  44. Study23 Group Title
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    Yea, notspecifically in this set of problems, but in a previous set it did

    • 2 years ago
  45. amistre64 Group Title
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    so, write up the derivate that we have so far, then the rest should become clear

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

    • 2 years ago
  47. amistre64 Group Title
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    and the given point tells us the values for x and y, so plug those in as well

    • 2 years ago
  48. Study23 Group Title
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    What about the y'? Do I need to isolate it??

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

    • 2 years ago
  50. amistre64 Group Title
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    and that y' IS the slope of thhe tangent line at the given point

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

    • 2 years ago
  52. amistre64 Group Title
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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?

    • 2 years ago
  53. Study23 Group Title
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    Slope-Point Formula! O R Or as it is more commonly known as, Point-Slope Form

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

    • 2 years ago
  55. Study23 Group Title
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    Isnt it -9/4? Not y'/4? What happened to that y?

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

    • 2 years ago
  57. Study23 Group Title
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    9/4

    • 2 years ago
  58. amistre64 Group Title
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    \[y=\frac94\]\[\frac29yy'=\frac29\frac94y'=\frac12y'\]

    • 2 years ago
  59. Study23 Group Title
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    Okay, that makes sense. That slope is correct btw per the book!

    • 2 years ago
  60. amistre64 Group Title
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    :)

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

    • 2 years ago
  62. amistre64 Group Title
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    my internet connection is toast, so ill have to say gnite :) good luck

    • 2 years ago
  63. Study23 Group Title
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    :D

    • 2 years ago
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