Question

Use implicit differentiation to find the equation of the tangent line to the curve xy^3+xy= 6 at the point ( 3 , 1 ). The equation of this tangent line can be written in the form y = mx+b where m is:
and where b is:

Answer 1

first you need \(y'\) so find the slope

Answer 2

ok

Answer 3

take the derivative using the chain rule and the product rule
you should get \[xy^3+xy= 6\]
\[y^3+3xy^2y'+y+xy'=0\]
ok with that? then we have to solve for \(y'\)

Answer 4

how do u have the slope?

Answer 5

we have nothing yet until we solve this equation for \(y'\)

Answer 6

o ok. i see.

Answer 7

you have a choice, you can plug in the numbers now for x and y, or you can use algebra to solve for \(y'\) first and then plug in the numbers. either way

Answer 8

ok

Answer 9

\[y^3+3xy^2y'+y+xy'=0\]
\[3xy^2y'+xy'=-y-y^3\]
\[y'(xy^2+x)=-y-y^3\]
\[y'=\frac{-y-y^3}{xy^2+x}\] now replace x by 3, y by 1 and find \(y'\) that is your slope.
then point- slope formula

Answer 10

ok let me try

Answer 11

ok let me know if you get stuck

Answer 12

slop is -2/9?

Answer 13

lets see
i get \(-\frac{2}{6}=-\frac{1}{3}\) but i could be wrong

Answer 14

ok let me see

Answer 15

no -1/3 is no correct as per my hw

Answer 16

yeah cause i am an idiot and dropped a 3 somewhere
you are right

Answer 17

:) im not totally hopeless in math then .lol

Answer 18

should be
\[y'=\frac{-y-y^3}{3xy^2+x}\]

Answer 19

should be minus right?

Answer 20

im not sure. i got -2/9. but my hw still said it was wrong

Answer 21

lets see if i can get a correct answer
\[\frac{-1-1}{3\times 3+3}=\frac{-2}{12}=-\frac{1}{6}\] who about that one?

Answer 22

1 + 9y' + 1 + 3y' = 0
12y' = -2
->y' = -2/12 = -16

Answer 23

ok thats it 1/-6

Answer 24

I mean -1/6

Answer 25

@Chlorophyll bless you !

Answer 26

yes thanks...

Answer 27

knew we would get there eventually

Answer 28

@satellite73 I always is the slowest one :")

Answer 29

yeah and i make all the mistakes!

Answer 30

3/2 is b

Answer 31

and i dont know what the hell im doing with math

Answer 32

now you are using the old point-slope formula
\[y-y_1=m(x-x_1)\] with
\[x_1=3,y_1=1,m=-\frac{1}{6}\]

Answer 33

\[y-1=-\frac{1}{6}(x-3)\] etc

Answer 34

@satellite73 If I spend time to explain like you does, I surely make mistakes at every single line :(