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

- anonymous

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

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- anonymous

i am thininking that
dx/dx 1 = dy/dx 1cosinx2y
would anyone like to tell me what they think?

- 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

zbay, there is no dx/dx. Find the derivative of each item. btw derivative of 1 is zero

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## More answers

- anonymous

right but isn't the derivative of x, 1?

- anonymous

Yeah, but don't put dx/dx it is confusing only when you find derivative of a y you include dy/dx

- 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

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

No you need to do product rule with the xsin(xy^2)

- anonymous

xsin(xy^2)+2cos(2)?

- anonymous

Hold on zbay. you have attempted a semi-complicated problem

- 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

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

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

Zbay, you might want to try a simpler problem and file this one for when you have your feet under you.

- 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

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

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

zbay what is this assignment or what course?

- anonymous

i need to find the dy/dx using implicit differentiation of the above listed equation

- anonymous

and it's in calc 1

- 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

Cal I with this difficult questions, has to be a take-home

- 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

OK, mine is wrong, let me double check

- 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

Yeah, go with zbay answer where he says Sorry a little messy.

- anonymous

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

- anonymous

Xavier is pretty good and it says he is just in high school He would be the smartest guy at my college.

- anonymous

I like math so i usually do self study. Doubt i'd be the smartest though.

- anonymous

So what to do study? Give me some pointers for my own self study.

- anonymous

this is a good video link. seems that it will be very helpfull to help follow along in class.

- anonymous

Oh yeah, PatrickJMT on youtube is good too.

- anonymous

I usually use a textbook for problems and use MIT OCW and Khan Academy

- anonymous

So how does that OCW work. Do you have to order something?

- anonymous

no it's a youtube video that put online and it's one of the luctures from mit

- anonymous

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

- anonymous

Thanks guys. Zbay if you get stuck shoot a question to me or xavier.

- anonymous

Also Single Variable Calc MIT OCW has its own openstudy page but it doesn't get as much love =(

- anonymous

hey thanks again for the help guys sure i will run into you guys again around here.

- anonymous

Good luck

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