Question

using implicit differentiation, find dy/dx of (x^2)y+x(y^2) = 6

Answer 1

The easiest way to do this is to use implicit differentiation on it as it stands.

You differentiate using the normal rules but you treat y as an unknown function of x. This means f(y) differentates to f '(y)*(dy/dx) by the chain rule.

xy + x + y^2 = 6
First implicit differentiation gives
[1*y + x*1*(dy/dx)] + 1 + 2y*(dy/dx) = 0
.....[from xy].....................[from y^2]
You can rearrange this to give dy/dx = (function of x and y)

Second implicit differentiation gives
dy/dx + [1*(dy/dx) + x*(d2y/dx2)] + [2*(dy/dx)*(dy/dx) + 2y*(d2y/dx2)] = 0
...................[from x*(dy/dx)]......................[from 2y*(dy/dx)]

Now you replace each dy/dx by its function of x and y found above.
Then rearrange it to give d2y/dx2 = (function of x and y).

Answer 2

i don't understand how you split it up, i'm confused.... that honestly made no sense

Answer 3

yay now my head hurts even worse

Answer 4

How Much More Questions Do You Have ?

Answer 5

Or Is This The Last Question ?

Answer 6

this is my first question...

Answer 7

How Does Your Head Hurt ?

Answer 8

i've got about 16 more for this section, then 5 on a second section on it
and a worksheet

Answer 9

i'm a junior in highschool taking calculus.. this isn't making sense

Answer 10

Oh Wow Genius :D

Answer 11

obviously not if this isn't making sense. sorry it's just frustrating!

Answer 12

and it's not making a bit of sense

Answer 13

Well, (x^2)y + x(y^2) = 6
d/dx[(x^2)y + x(y^2)] = d/dx (6)

Answer 14

2xy + dy/dx(x^2) + y^2 + 2y(dy/dx)x =0

Answer 15

i'm lost stop

Answer 16

how'd you get 2xy + dy/dx(x^2 + y^2 + 2y(dydx)x=0?

Answer 17

You dont know how to differentiate ?

Answer 18

I do, but it's confusing me

Answer 19

the d/dx of x^2y?

Answer 20

you use the multiplication rule right? udv + vdu?

Answer 21

\[x ^{2}y+xy^2=6\]

Answer 22

x^2(1) + y(2x) = 2xy + x^2

Answer 23

Yes I am Using Product Rule

Answer 24

then how did you get 2xy then dy/dx x^2?

Answer 25

or do you get dy/dx and all that stuff in parenthasis because you're finding the d/dx of y

Answer 26

\[\frac{ dy }{ dx }[(x^2)y + x(y^2)] =\frac{ dy }{ dx }6\]

Answer 27

Now Use Product Rule

Answer 28

\[2xy + dy/dx(x^2) + y^2 + 2y(dy/dx)x =0\]

Answer 29

Now Take The Terms Including dy/dx On One Side

Answer 30

\[dy/dx(x^2)+2y(dy/dx)x = -y^2-2xy\]

Answer 31

Now Take Common dy/dx

Answer 32

\[dy/dx(x^2+2xy)=-y^2-2xy\]

Answer 33

then divide?

Answer 34

So,\[\frac{ dy }{ dx }=\frac{ -(y^2+2xy)}{x^2+2xy }\]

Answer 35

^ Now Thats Your Answer Got It :)

Answer 36

question, when using the product rule

Answer 37

how do you know which one to put dy/dx with

Answer 38

like when we had x^2y and got dy/dx(x^2) + 2xy

Answer 39

why'd we put dy/dx with x^2 and not 2xy?

Answer 40

Well Its Your Own Decision Whether to Put U first or V First If We talk about the formula

Answer 41

okay but then which one do i put the dy/dx with?

Answer 42

like the next problem is y2 = (x+1)/(x-1)

Answer 43

so would i do 2y (dy/dx) = vdu - udv/v^2

Answer 44

Depends Completely On You

Answer 45

so dy/dx2y = (x+1)(1) - (x-1)(1) / (x-1)^2

Answer 46

@Godsgirl Please Help Her Out I Gotta Leave- Thanks :)

Answer 47

Hey so wht do u need help with

Answer 48

Hello?

Answer 49

I'm here. I have a question on where to put the dy/dx when solving

Answer 50

right now i'm working on y^2 = x-1/x+1

Answer 51

so i take dy/dx y^2 = (x+1)(1) - (x-1)(1) / (x+1)^2 right?

Answer 52

right

Answer 53

and the left side will equal 2y when i take dy/dx of it

Answer 54

now where do i keep dy/dx, when solving?

Answer 55

here i already solved it

Answer 56

where do i put it? because i know i need to get it alone

Answer 57

I sent it to u on ur messages

Answer 58

that's not the question i'm asking about..

Answer 59

the new one is

Answer 60

which one are u asking about then

Answer 61

y^2 = x-1/x+1

Answer 62

i'm not trying to come off as rude, i'm sorry!

Answer 63

just frustrated

Answer 64

I am Back

Answer 65

\[y^2 = (x-1) /(x+1)\]

Answer 66

differentiate both sides with respect to x

Answer 67

oh okay i was lost there

Answer 68

i see where shes at now

Answer 69

\[2y dy/dx = [(x+1)-(x-1)] /(x+1)^2 \]

Answer 70

\[2y dy/dx = 2/(x+1)^2\]

Answer 71

\[dy/dx = y / (x+1)^2\]

Answer 72

@jamroz Got It :)

Answer 73

so you keep dy/dx with the y on the left

Answer 74

but why? cause i had gotten that far, i just didn't know which side to leave dy/dx with?

Answer 75

Now You Got Which Side You Had To Keep It :) Anyways I am Off To Sleep, Aww My Head Hurts Anyways Bye :)

Answer 76

thank you so much!