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Study23

  • 2 years ago

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'???

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  1. Study23
    • 2 years ago
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    Im referring to implicit differentiation...

  2. amistre64
    • 2 years ago
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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

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

  4. Study23
    • 2 years ago
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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'?

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

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

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

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

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

  10. amistre64
    • 2 years ago
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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

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

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

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

  14. Study23
    • 2 years ago
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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?

  15. amistre64
    • 2 years ago
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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

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

  17. amistre64
    • 2 years ago
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    correct

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

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

  20. amistre64
    • 2 years ago
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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

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

  22. Study23
    • 2 years ago
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    I see.

  23. amistre64
    • 2 years ago
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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

  24. Study23
    • 2 years ago
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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?

  25. amistre64
    • 2 years ago
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    sure

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

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

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

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

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

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

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

  33. Study23
    • 2 years ago
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    I see now

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

  35. amistre64
    • 2 years ago
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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

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

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

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

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

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

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

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

  43. amistre64
    • 2 years ago
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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

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

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

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

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

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

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

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

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

  52. amistre64
    • 2 years ago
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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?

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

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

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

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

  57. Study23
    • 2 years ago
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    9/4

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

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

  60. amistre64
    • 2 years ago
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    :)

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

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

  63. Study23
    • 2 years ago
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    :D

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