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i am thininking that dx/dx 1 = dy/dx 1cosinx2y would anyone like to tell me what they think?
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?
zbay, there is no dx/dx. Find the derivative of each item. btw derivative of 1 is zero
right but isn't the derivative of x, 1?
Yeah, but don't put dx/dx it is confusing only when you find derivative of a y you include dy/dx
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?
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.)
No you need to do product rule with the xsin(xy^2)
Hold on zbay. you have attempted a semi-complicated problem
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.
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.
The right hand side of eq should read cos x sin (xy^2) - (y^2 + 2xy dy/dx)cos (xy^2) sin (xy^2)
Zbay, you might want to try a simpler problem and file this one for when you have your feet under you.
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
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
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)
zbay what is this assignment or what course?
i need to find the dy/dx using implicit differentiation of the above listed equation
and it's in calc 1
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
Cal I with this difficult questions, has to be a take-home
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?
OK, mine is wrong, let me double check
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
Yeah, go with zbay answer where he says Sorry a little messy.
To keep up just watch the MIT Single variable calculus course. http://www.youtube.com/watch?v=7K1sB05pE0A&playnext=1&list=PL590CCC2BC5AF3BC1
Xavier is pretty good and it says he is just in high school He would be the smartest guy at my college.
I like math so i usually do self study. Doubt i'd be the smartest though.
So what to do study? Give me some pointers for my own self study.
this is a good video link. seems that it will be very helpfull to help follow along in class.
Oh yeah, PatrickJMT on youtube is good too.
I usually use a textbook for problems and use MIT OCW and Khan Academy
So how does that OCW work. Do you have to order something?
no it's a youtube video that put online and it's one of the luctures from mit
Nah OCW is opencouseware. Its videos and course materials they offer on the web for free. http://ocw.mit.edu/index.htm
Thanks guys. Zbay if you get stuck shoot a question to me or xavier.
Also Single Variable Calc MIT OCW has its own openstudy page but it doesn't get as much love =(
hey thanks again for the help guys sure i will run into you guys again around here.