Question

determine the slope of the following curve at the given point.
6xy^(1/2)+y=13; (2,1)

Answer 1

major help needed

Answer 2

\[6x\sqrt{y}+y=13\] right?

Answer 3

written a diff way but yes

Answer 4

so your job it to find y' using implicit differentiation and evaluate at (2,1)

Answer 5

the derivative of y is no problem, it is y'
and the derivative of 13 is 0
the work is in finding the derivative of
\[6x\sqrt{y}\] which requires the product rule

Answer 6

slope is def not 1

Answer 7

-6/7

Answer 8

so taking the derivative wrt x you get
\[6\sqrt{y}+6x\times \frac{1}{2\sqrt{y}}\times y'\]

Answer 9

that is the only complicated step in this problem
knowing to use the produce rule and the chain rule. this gives
\[6\sqrt{y}+\frac{3y'}{\sqrt{y}}+y'=0\]

Answer 10

ok

Answer 11

i made a typo hold on

Answer 12

\[6\sqrt{y}+\frac{3xy'}{\sqrt{y}}+y'=0\]

Answer 13

i forgot the x.
now you can either use algebra now to solve for y' and then replace x by 2 and y by 1
or else you can replace x by 2, y by 1 and then solve for y'
either way

Answer 14

i just need the slope though

Answer 15

yeah, is it clear what you have to do?

Answer 16

you want y' when x = 2 and y = 1

Answer 17

do you understand the whole equation on a tangent line
that concept..i do not understand..ill give you an example

Answer 18

\[6\sqrt{y}+\frac{3xy'}{\sqrt{y}}+y'=0\] so if x = 2, y = 1 you get
\[6\sqrt{1}+\frac{3\times 2y'}{\sqrt{1}}+y'=0\] or just
\[6+6y'+y'=0\] and solve this for y'

Answer 19

you get
\[y'=-\frac{6}{7}\]

Answer 20

like determine 2x^2+3xy+3y^2=17;(2,1)

Answer 21

i got that =)

Answer 22

right, use implicit diff, then you will get y' in terms of x and y, then replace x and y by the numbers given and you will get the slope

Answer 23

i always seem to make stupid mistakes

Answer 24

so in your example step one would be
\[4x+3y+3xy'+6yy'=0\] and then either solve for y' directly and substitute, or substitute and solve for y'

Answer 25

ok..ill try again =(

Answer 26

its just frustrating lol

Answer 27

important part is to remember product and chain rule, and also that you are thinking
\[y=f(x), y'=f'(x)\] so for example if you see
\[\sqrt{x}y\] you think
\[\sqrt{x}f(x)\] and if you see
\[xy^2\] you think
\[xf^2(x)\] that will remind you to use product and chain rule

Answer 28

did you get -4/3x - 11/12?

Answer 29

thats what i got

Answer 30

its wrong =(

Answer 31

help?

Answer 32

??