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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am thininking that dx/dx 1 = dy/dx 1cosinx2y would anyone like to tell me what they think?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Don'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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0zbay, there is no dx/dx. Find the derivative of each item. btw derivative of 1 is zero

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right but isn't the derivative of x, 1?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, but don't put dx/dx it is confusing only when you find derivative of a y you include dy/dx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Don'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.)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No you need to do product rule with the xsin(xy^2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hold on zbay. you have attempted a semicomplicated problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The right hand side of eq should read cos x sin (xy^2)  (y^2 + 2xy dy/dx)cos (xy^2) sin (xy^2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Zbay, you might want to try a simpler problem and file this one for when you have your feet under you.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha, i hear you chaguanas however the level of difficulty seems to be the same accross the board for all the problems in this assignment

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Xavier, 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)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0zbay what is this assignment or what course?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i need to find the dy/dx using implicit differentiation of the above listed equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Cal I with this difficult questions, has to be a takehome

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK, mine is wrong, let me double check

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, go with zbay answer where he says Sorry a little messy.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0To keep up just watch the MIT Single variable calculus course. http://www.youtube.com/watch?v=7K1sB05pE0A&playnext=1&list=PL590CCC2BC5AF3BC1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Xavier is pretty good and it says he is just in high school He would be the smartest guy at my college.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I like math so i usually do self study. Doubt i'd be the smartest though.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So what to do study? Give me some pointers for my own self study.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is a good video link. seems that it will be very helpfull to help follow along in class.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh yeah, PatrickJMT on youtube is good too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I usually use a textbook for problems and use MIT OCW and Khan Academy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So how does that OCW work. Do you have to order something?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no it's a youtube video that put online and it's one of the luctures from mit

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Nah OCW is opencouseware. Its videos and course materials they offer on the web for free. http://ocw.mit.edu/index.htm

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks guys. Zbay if you get stuck shoot a question to me or xavier.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Also Single Variable Calc MIT OCW has its own openstudy page but it doesn't get as much love =(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey thanks again for the help guys sure i will run into you guys again around here.
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