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anonymous

  • 5 years ago

alright need to find the dy/dx by implicit differentiation of 1+x=xsin(xy^2)

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  1. anonymous
    • 5 years ago
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    i am thininking that dx/dx 1 = dy/dx 1cosinx2y would anyone like to tell me what they think?

  2. anonymous
    • 5 years ago
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    Don't you need to product rule first. Then when differentiating sin(xy^2) do chain rule and then do product rule again for xy^2?

  3. anonymous
    • 5 years ago
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    zbay, there is no dx/dx. Find the derivative of each item. btw derivative of 1 is zero

  4. anonymous
    • 5 years ago
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    right but isn't the derivative of x, 1?

  5. anonymous
    • 5 years ago
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    Yeah, but don't put dx/dx it is confusing only when you find derivative of a y you include dy/dx

  6. anonymous
    • 5 years ago
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    i am going to have to go back and read more about implicit differentiation, but i thought it was used to compare the change of derivitive in y compared to change in derivative from x? Other than that did i derive at the correct answer on the right side with the trig portion of the function?

  7. anonymous
    • 5 years ago
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    Don't read too much in to it. Implicit differentiation simply means your y is imbedded in the equation, rather than y=x^2. (Checking on the other side.)

  8. anonymous
    • 5 years ago
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    No you need to do product rule with the xsin(xy^2)

  9. anonymous
    • 5 years ago
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    xsin(xy^2)+2cos(2)?

  10. anonymous
    • 5 years ago
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    Hold on zbay. you have attempted a semi-complicated problem

  11. anonymous
    • 5 years ago
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    You also need to do chain rule with the sin(xy^2) (uv)'=u'v+uv' sin(xy^2)+xcos(xy^2)(y^2+2xyy') The second part inside the parenthesis as the end is the chain rule on the insides of sin(xy^2) again using the product rule.

  12. anonymous
    • 5 years ago
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    ok this may seem like a dumb question but wouldn't the "x" in the xcos become a 1 and therefore not be included in the equation? i am just wondering i am tracking with the rest you said to do.

  13. anonymous
    • 5 years ago
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    The right hand side of eq should read cos x sin (xy^2) - (y^2 + 2xy dy/dx)cos (xy^2) sin (xy^2)

  14. anonymous
    • 5 years ago
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    Zbay, you might want to try a simpler problem and file this one for when you have your feet under you.

  15. anonymous
    • 5 years ago
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    haha, i hear you chaguanas however the level of difficulty seems to be the same accross the board for all the problems in this assignment

  16. anonymous
    • 5 years ago
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    That entire thing is the differentiation of the right hand side. xsin(xy^2) right So i use the product rule d/dx(x)*sin(xy^2)+x*d/dx(sin(xy^2)) 1*sin(xy^2)+x*cos(xy^2) BUT i have to use the chain rule on what is inside the cosine. This requires the product rule again 1*sin(xy^2)+x*cos(xy^2)(1*y^2+x2y(dy/dx)) and so you have sin(xy^2)+xcos(xy^2)(y^2+2xy(dy/dx)) for the right side Sorry its a bit messy

  17. anonymous
    • 5 years ago
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    Xavier, you are missing something. You put 1 as your u' but u' should be cos xy^2. (Above in my answer I put cos x but of course it should be cos xy^2)

  18. anonymous
    • 5 years ago
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    zbay what is this assignment or what course?

  19. anonymous
    • 5 years ago
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    i need to find the dy/dx using implicit differentiation of the above listed equation

  20. anonymous
    • 5 years ago
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    and it's in calc 1

  21. anonymous
    • 5 years ago
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    Yes, I gave my answer a few posts up. I think it is right, I see one or two mistakes with xavier like a plus sign should be a minus. Put the whole differentiated eq together and solve for dy/dx

  22. anonymous
    • 5 years ago
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    Cal I with this difficult questions, has to be a take-home

  23. anonymous
    • 5 years ago
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    The one is the derivative of x. u=x v=sin(xy^2) (uv)'=u'v+uv' d/dx(x)*sin(xy^2)+x*d/dx(sin(xy^2)) Is what i did Which mistakes?

  24. anonymous
    • 5 years ago
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    OK, mine is wrong, let me double check

  25. anonymous
    • 5 years ago
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    thanks for the help guys it's helping understand this stuff a bit better. I kinda screwed myself skiping trig and going to calc thinking that i could cut down on the amount of courses i would have to take

  26. anonymous
    • 5 years ago
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    Yeah, go with zbay answer where he says Sorry a little messy.

  27. anonymous
    • 5 years ago
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    To keep up just watch the MIT Single variable calculus course. http://www.youtube.com/watch?v=7K1sB05pE0A&playnext=1&list=PL590CCC2BC5AF3BC1

  28. anonymous
    • 5 years ago
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    Xavier is pretty good and it says he is just in high school He would be the smartest guy at my college.

  29. anonymous
    • 5 years ago
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    I like math so i usually do self study. Doubt i'd be the smartest though.

  30. anonymous
    • 5 years ago
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    So what to do study? Give me some pointers for my own self study.

  31. anonymous
    • 5 years ago
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    this is a good video link. seems that it will be very helpfull to help follow along in class.

  32. anonymous
    • 5 years ago
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    Oh yeah, PatrickJMT on youtube is good too.

  33. anonymous
    • 5 years ago
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    I usually use a textbook for problems and use MIT OCW and Khan Academy

  34. anonymous
    • 5 years ago
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    So how does that OCW work. Do you have to order something?

  35. anonymous
    • 5 years ago
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    no it's a youtube video that put online and it's one of the luctures from mit

  36. anonymous
    • 5 years ago
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    Nah OCW is opencouseware. Its videos and course materials they offer on the web for free. http://ocw.mit.edu/index.htm

  37. anonymous
    • 5 years ago
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    Thanks guys. Zbay if you get stuck shoot a question to me or xavier.

  38. anonymous
    • 5 years ago
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    Also Single Variable Calc MIT OCW has its own openstudy page but it doesn't get as much love =(

  39. anonymous
    • 5 years ago
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    hey thanks again for the help guys sure i will run into you guys again around here.

  40. anonymous
    • 5 years ago
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    Good luck

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