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mathcalculus

  • one year ago

HELP Please. Use implicit differentiation to find the equation of the tangent line to the curve xy^3+xy=4 at the point (2,1) . The equation of this tangent line can be written in the form: y=mx+b where M is: and where B is:

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  1. dumbcow
    • one year ago
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    remember product rule for implicit \[(fg)' = f'g +fg'\]

  2. dumbcow
    • one year ago
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    M = slope = dy/dx

  3. mathcalculus
    • one year ago
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    oh so for implicit we use the product?

  4. mathcalculus
    • one year ago
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    always?

  5. dumbcow
    • one year ago
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    yes and chain rule....since it is with respect to x \[d/dx f(y) = f'(y)*\frac{dy}{dx}\]

  6. mathcalculus
    • one year ago
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    so which is f and which is g

  7. mathcalculus
    • one year ago
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    im a little confused since xy^3 are together.

  8. dumbcow
    • one year ago
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    depends on the term...first there is "xy^3" f(x) = x g(y) = y^3

  9. mathcalculus
    • one year ago
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    so f = x and g= y^3 ?

  10. mathcalculus
    • one year ago
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    so is the derivative: x*3y+ y^3*1 ?

  11. dumbcow
    • one year ago
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    yes, so result is \[1*y^{3}+ x*3y^{2} \frac{dy}{dx}\]

  12. mathcalculus
    • one year ago
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    dy/dx?

  13. dumbcow
    • one year ago
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    thats what the slope is...which is what we are trying to find

  14. mathcalculus
    • one year ago
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    all i know is the product rule is this: \[\frac{ d }{ dx } (fg)= f g \prime+g f \prime\]

  15. dumbcow
    • one year ago
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    anytime you differentiate a function of "y" wrt x you must multiply by "dy/dx" refer to my prev post

  16. mathcalculus
    • one year ago
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    otay. so then what.

  17. dumbcow
    • one year ago
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    repeat process for all terms of equation

  18. mathcalculus
    • one year ago
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    ?

  19. mathcalculus
    • one year ago
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    what do you mean

  20. dumbcow
    • one year ago
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    xy^3 +xy = 4 you have to differentiate each term, the "xy^3" and "xy" adn "4"

  21. mathcalculus
    • one year ago
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    3xy^2.....

  22. dumbcow
    • one year ago
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    ?? first term is \[(y^{3} +3xy^{2} \frac{dy}{dx})\]

  23. dumbcow
    • one year ago
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    next term is "xy" .... derivative is \[(y + x \frac{dy}{dx})\] then of course , derivative of constant is 0 now you have \[y^{3} +3xy^{2} \frac{dy}{dx} +y+x \frac{dy}{dx} = 0\]

  24. dumbcow
    • one year ago
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    solve for "dy/dx" in terms of x and y

  25. mathcalculus
    • one year ago
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    okay i understand from here: \[\frac{ dy }{ dx}= 3xy ^{2}+ y ^{3}\]

  26. mathcalculus
    • one year ago
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    then from there, i don't know what to do.

  27. mathcalculus
    • one year ago
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    @dumbcow

  28. dumbcow
    • one year ago
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    well from there you would plug in the given point (2,1) to give you the slope (M) of tangent line but how did you solve for dy/dx .... your answer is wrong

  29. mathcalculus
    • one year ago
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    i didn't i'm doing this step by step

  30. mathcalculus
    • one year ago
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    from the result.. ^^ i just multiplied.

  31. dumbcow
    • one year ago
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    ?? sorry i am not understanding

  32. mathcalculus
    • one year ago
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    remember before you wrote "yes, result is bla blah" then from there i just multiplied... x*(3y)^2 + y^3 *1

  33. mathcalculus
    • one year ago
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    you know.

  34. dan815
    • one year ago
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    hey mathcalculus

  35. dan815
    • one year ago
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    do you know chain rule

  36. dan815
    • one year ago
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    |dw:1363334172119:dw|

  37. dumbcow
    • one year ago
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    let me just show the steps, maybe that will help \[y^{3}+3xy^{2} \frac{dy}{dx} +y+x \frac{dy}{dx} = 0\] \[3xy^{2} \frac{dy}{dx}+x \frac{dy}{dx} = -y^{3}-y\] \[\frac{dy}{dx}(3xy^{2}+x) = -y^{3}-y\] \[\frac{dy}{dx} = \frac{-y^{3}-y}{3xy^{2}+x}\]

  38. dumbcow
    • one year ago
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    i get the feeling i lost you when i started using "dy/dx" :{

  39. mathcalculus
    • one year ago
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    mhm :/

  40. mathcalculus
    • one year ago
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    |dw:1363334405957:dw| @dan815

  41. dan815
    • one year ago
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    |dw:1363334256950:dw|

  42. dan815
    • one year ago
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    think about it like this Y is a function of X right so say you have to take the derivative of 2x^2 what is it?

  43. dan815
    • one year ago
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    |dw:1363334523512:dw|

  44. mathcalculus
    • one year ago
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    wait wait wait

  45. dumbcow
    • one year ago
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    thx @dan815 for clarifying :) i didn't explain very well

  46. dan815
    • one year ago
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    remember how chain rule works

  47. mathcalculus
    • one year ago
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    the chain rule: if f(x)= g(k(x)) then f prime x= g prime (k(x))k prime (x) correct?

  48. mathcalculus
    • one year ago
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    @dumbcow you explained very good too.

  49. dan815
    • one year ago
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    @dumbcow no its okay lol i can relate to him cuz i know how noobs think :)

  50. mathcalculus
    • one year ago
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    i just can't understand @dan815 techwriting lol

  51. dan815
    • one year ago
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    yes that is chain rule

  52. mathcalculus
    • one year ago
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    noob? -_-

  53. mathcalculus
    • one year ago
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    411- not a guy.

  54. dan815
    • one year ago
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    haha trust me we are all noobs, just wait till you see how much those phd profs know

  55. mathcalculus
    • one year ago
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    okay dc back to math.

  56. mathcalculus
    • one year ago
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    thank god, no more crazy math after this. but seriously back to work lol

  57. mathcalculus
    • one year ago
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    i have to understand this.

  58. dan815
    • one year ago
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    ok ok so for implicit differentiation when you see Y think about it like some function on x

  59. mathcalculus
    • one year ago
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    ello?

  60. mathcalculus
    • one year ago
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    yup got that.

  61. mathcalculus
    • one year ago
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    then?

  62. dan815
    • one year ago
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    ok do this

  63. mathcalculus
    • one year ago
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    so first i was told to use the product rule.. then the chain. which that confused me a little. so lets take it step by step please.

  64. dan815
    • one year ago
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    |dw:1363334909090:dw|

  65. dan815
    • one year ago
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    yes you are going to need both those rules for this

  66. dan815
    • one year ago
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    no remember than y is like a function of x

  67. mathcalculus
    • one year ago
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    OH

  68. mathcalculus
    • one year ago
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    darn it.

  69. dan815
    • one year ago
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    |dw:1363334981033:dw|

  70. mathcalculus
    • one year ago
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    so we use the product rule 1st.

  71. dan815
    • one year ago
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    sub 2x where you see Y and solve it

  72. mathcalculus
    • one year ago
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    i have no idea what im doing.

  73. mathcalculus
    • one year ago
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    for some reason I'm all confused.

  74. dan815
    • one year ago
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    okay look at this question again same question but im telling you what y is

  75. mathcalculus
    • one year ago
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    don't tell me what y is. how am i suppose to know alone?

  76. mathcalculus
    • one year ago
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    let's do that problem that I gave.

  77. dan815
    • one year ago
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    |dw:1363335163468:dw|

  78. dan815
    • one year ago
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    just do this im trying to teach you how to do the problem above and every other problem like this

  79. mathcalculus
    • one year ago
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    k -.-

  80. mathcalculus
    • one year ago
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    x= 2/5 ?

  81. mathcalculus
    • one year ago
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    i plugged in 2x..

  82. dan815
    • one year ago
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    wut no find dervivative

  83. dan815
    • one year ago
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    chain rule

  84. mathcalculus
    • one year ago
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    ?

  85. dan815
    • one year ago
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    |dw:1363335349192:dw|

  86. dan815
    • one year ago
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    do you get how i got that when i differentiated there

  87. dan815
    • one year ago
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    whats confusing you there is it the d/dx?

  88. dan815
    • one year ago
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    i know that you know how to take derivate of Xs

  89. dan815
    • one year ago
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    you there?

  90. dumbcow
    • one year ago
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    any progress being made here...

  91. mathcalculus
    • one year ago
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    yeah im looking.

  92. dan815
    • one year ago
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    ya i think so lol we've found the root problem here

  93. dan815
    • one year ago
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    she just doesnt know what it means to difference functions as opposed to variables

  94. mathcalculus
    • one year ago
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    @dumbcow you used the product rule first which i was understanding.... until mr. CHAIN RULE came along to make a knot.

  95. mathcalculus
    • one year ago
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    :)

  96. dan815
    • one year ago
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    |dw:1363335696884:dw|

  97. dan815
    • one year ago
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    do you understand this yet?

  98. dumbcow
    • one year ago
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    yeah...sorry i figured that was implied since we are dealing with implicit differentiation :|

  99. mathcalculus
    • one year ago
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    yeah @dan815 but now why did you multiply 2 at the end.?

  100. dan815
    • one year ago
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    thats chain rule remember you have to multiply by the derivative of the function inside the bracket

  101. mathcalculus
    • one year ago
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    i understand you got the derivative

  102. mathcalculus
    • one year ago
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    |dw:1363335825708:dw|

  103. dan815
    • one year ago
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    |dw:1363335821148:dw|

  104. dan815
    • one year ago
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    this is known as the chain rule

  105. mathcalculus
    • one year ago
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    okie let's do our problem @dan815

  106. mathcalculus
    • one year ago
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    i know youre trying to teach me, so i can do the rest.. etc; but i need to do one problem first in order to proceed.

  107. mathcalculus
    • one year ago
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    i remember the chain rule a little better now.

  108. dan815
    • one year ago
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    so if its |dw:1363335898535:dw|

  109. dan815
    • one year ago
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    ok lets go back to our problem knowing the chain rule now

  110. mathcalculus
    • one year ago
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    gratzie

  111. dan815
    • one year ago
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    what i really wanted you to undetsand was this concept though basically you have to think of Y as a function so when you see Ys differentiating Y is like differentiation of a function so think chain rules

  112. dan815
    • one year ago
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    |dw:1363336027415:dw|

  113. dan815
    • one year ago
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    so for the implict up there

  114. mathcalculus
    • one year ago
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    ok

  115. mathcalculus
    • one year ago
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    got it, think the x as in y.

  116. mathcalculus
    • one year ago
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    so our problem... is this: xy^3+xy=4 at the point (2,1) .

  117. mathcalculus
    • one year ago
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    start by:

  118. dan815
    • one year ago
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    |dw:1363336084849:dw|

  119. dan815
    • one year ago
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    |dw:1363336247542:dw|