Question

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

Answer 1

i am thininking that
dx/dx 1 = dy/dx 1cosinx2y

would anyone like to tell me what they think?

Answer 2

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?

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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?

Answer 7

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.)

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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.

Answer 12

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.

Answer 13

The right hand side of eq should read
cos x sin (xy^2) - (y^2 + 2xy dy/dx)cos (xy^2) sin (xy^2)

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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)

Answer 18

zbay what is this assignment or what course?

Answer 19

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

Answer 20

and it's in calc 1

Answer 21

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

Answer 22

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

Answer 23

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?

Answer 24

OK, mine is wrong, let me double check

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

Oh yeah, PatrickJMT on youtube is good too.

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

Good luck