Question

Find dy/dx.
2x^2y −3/y = 0. How do I do this one?

Answer 1

\[\Large\rm 2x^2y-\frac{3}{y}=0\]Is this the correct formatting?

Answer 2

yes it is :)

Answer 3

\[\Large\rm 2x^2y-3y^{-1}=0\]So looks like we need to start with product rule, yes?\[\Large\rm derivative:\qquad \color{royalblue}{(2x^2)'}y+2x^2\color{royalblue}{(y)'}-\color{royalblue}{(3y^{-1})'}=0\]Blue parts are where we need to take a derivative.

Answer 4

ok so 1st one derivative is : 4x
2nd one is:1
3rd one is: .... -3?

Answer 5

The 2nd one is 1y'

Answer 6

or just y'

Answer 7

...

Answer 8

Then not sure how to step forward from here.. Product rule?

Answer 9

@zepdrix is the answer 4xy + 2x^2 - 3/y^2?

Answer 10

Example of product rule, and implicit differentiation:
\((xy^2)'=x'y^2+x(y^2)'=1*y^2+x(2y*y')=y^2+2xyy'\)

Answer 11

isn't it easier to use quotient rule for 3/y? it becomes 0.y - 3.1 divided by y^2 or - 3/y^2?

Answer 12

@kennybm
Note the y' has to be attached after if you use the chain rule to differentiate with respect to y firrst:
\((y^2)'=2yy'\)

Answer 13

I'm so lost right now... I'm trying to take this course online and it hasn't explained all this stuff to me..

Answer 14

I will give a medal to whoever can help me...

Answer 15

and a good testimony.. Somebody, please???

Answer 16

Do you follow @zepdrix 's first and second posts?

Answer 17

I do and I answered the first thing to do with finding derivative but what do I do now?

Answer 18

I got 4x, y and then -3

Answer 19

But they are not correct, and he was probably no longer on line.

Answer 20

Can we start from his second post?

Answer 21

Can you evaluate \((2x^2)y\)?

Answer 22

4x is not wrong, but not complete! :(

Answer 23

4x^1?

Answer 24

4xy

Answer 25

Yes, but you have a "y" stuck to it, so you have to keep it to make 4xy, right?
That's part of the product rule.

Answer 26

Yes, 4xy is right!
Now the second term!

Answer 27

do i take derivative of 2x^2? or just the y?

Answer 28

just the y!

Answer 29

just y would stay y right?

Answer 30

It's a "trick" question.
Since you don't know much about y, the derivative of y is y' ! :)

Answer 31

So the second term becomes...

Answer 32

@zydic you there?

Answer 33

yeah, sorry, so it is now 4xy, y and then next etrm, is it -3 or how do I do that?

Answer 34

The second term is not y.
2x^2(y')
Recall that in the product rule, the part that you don't differentiate will still be multiplied with the part that you differentiate.

Answer 35

oooooooh

Answer 36

so it would be 2x^2y

Answer 37

2x^2y means you did not differentiate anything.
You're supposed to differentiate y to get y'.
So the result is 2x^2y'.

Answer 38

oh, forgot the y' instead of y part..

Answer 39

ok so far before doing the third term?

Answer 40

so I would have 4xy+2x^2y'

Answer 41

Yes, plus the third term.

Answer 42

would i carry the - before the ( )into the equation?

Answer 43

It's up to you, but the result should be the same!

Answer 44

ok so would it be? -3/y^2

Answer 45

Almost, only you have to include the -ve sign before the parentheses.
(3/y)'=-3/y^2*y'
so
-(3/y)'=+3/y^2*y'
Do not forget that you need to use the chain rule.

Answer 46

d(K)/dx = d(K)/dy * (dy/dx)

Answer 47

where dy/dx = y'

Answer 48

....so i have 3/y^2*-3/y^2?

Answer 49

The third term is
+3/y^2*y'
The y' comes from the chain rule. +3/y^2 was obtained by differentiating with respect to y, but we are differentiating with respect to x.
Recall the chain rule
d(K)/dx = d(K)/dy * (dy/dx) = d(K)/dy * y'

Answer 50

ok, ok.. I'm still catching on here.. :)

Answer 51

−(3/y)′ = -d(3/y)/dy * dy/dx = -(-3/y^2) * y' = +(3/y^2)y'
Would this be better?

Answer 52

So it is such a long way but I am trying. so it is all together 4xy+2x^2y'++3/y^2*y'

Answer 53

ok, i see it :)

Answer 54

Good, now we do not forget the right hand side:
4xy+2x^2y'+(3/y^2)y' = 0

Answer 55

All you need to do now is to isolate y' (which means dy/dx) and solve for y'.
Can you do that?

Answer 56

oooh....

Answer 57

can you really differentiate a 0?

Answer 58

0 is a constant, so the derivative of a constant is...

Answer 59

0

Answer 60

Yep! You're doing great!

Answer 61

thanks, it's alot to take in... I appreciate all your patience with me.. and I appreciate all your help!

Answer 62

You're welcome! :)

Answer 63

Calculus is a tough subject! so, since the right hand side is 0 do I do anything more?

Answer 64

Since the derivative of 0 is 0, you end up with the equation:
4xy+2x^2y'+(3/y^2)y' = 0
from which you can solve for y' (=dy/dx), which is answer requested.

Answer 65

are they looking for the 0 or the whole thing written up? or do I have to take more derivatives?

Answer 66

They are looking for the derivative dy/dx, which equals y'.
By solving the equation for y', you will have the correct answer.

Answer 67

isn't this a quadratic?

Answer 68

It is more complex than a quadratic in x and y, but it is linear in y'.
So the solution is relative easy to find.

Answer 69

ohhh, geez, I'm not sure how to solve this then..

Answer 70

I'm getting more paper out...

Answer 71

4xy+2x^2y'+(3/y^2)y' = 0
2x^2y'+(3/y^2)y' = -4xy
y'(2x^2+3/y^2) = -4xy

Can you continue to find y'?

Answer 72

.is the next part y'=-4xy/2x^2+s/y^2?

Answer 73

oops that bottom s is supposed to be a 3/y^2

Answer 74

can't i somehow simplify the -4xy/2x^2+3/y^2?

Answer 75

You do not need to.

Answer 76

so that should be the final on what they are looking for?

Answer 77

The answer will contain a "mixture" of x and y (but not y').

Answer 78

y'(2x^2+3/y^2) = -4xy
=>
dy/dx= y' = -4xy/(2x^2+3/y^2)

Answer 79

y' is -4xy/(2x^2+3/y^2)

Answer 80

That is correct!

Answer 81

oh my goodness! That was such a complicated process! i hope I can get this before my midterm next week!! I've now got a headache! ;) wow, Thank you!!!! I'm now going to give you that wonderful testimony! :)

Answer 82

Rest a little before you continue! You're doing great!
It always pay to understand every step, or else you get stuck with the next problem.
:)

Answer 83

* pays

Answer 84

:)