anonymous
  • anonymous
alright need to find the dy/dx by implicit differentiation of 1+x=xsin(xy^2)
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
i am thininking that dx/dx 1 = dy/dx 1cosinx2y would anyone like to tell me what they think?
anonymous
  • anonymous
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?
anonymous
  • anonymous
zbay, there is no dx/dx. Find the derivative of each item. btw derivative of 1 is zero

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

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