Question

how do you find the derivative of (y+4)/(x-2)=5y at (3,1)???

Answer 1

either solve for "y" all by its little lonesome; or use implicit deriv..different.....that one word...

Answer 2

lets try implicit...

5y = (y+4)/(x-2) ; take the d/dx of both sides..

(d/dx)(5y) = (d/dx)[(y+4)/(x-2)]

Answer 3

5y' = apply quotient rule to this side

5y' = [(x-2)(1y') - (y+4)(x')] / [(x-2)^2] ; dy/dx = y' ; x' = 1

Answer 4

get y' by itself if you dare :)

Answer 5

no easier way? how can you get y by it self

Answer 6

lol....

I get:

y+4
y' = ------
x-7

Answer 7

how??

Answer 8

plug in (3,1)

y' = -5/6 ??

where did I lose ya? :)

Answer 9

y' = -5/4 ; forgot how to add lol

Answer 10

how did you get to y+4/x-7

Answer 11

by using paper and pencil really... but lets see if I can type it out...

Answer 12

(x-2)(1y') - (y+4)(x')
5y' = ---------------- ; move (x-2)^2 up and bring y' down..
(x-2)^2


(x-2)(y') - (y+4)
5(x-2)^2 = ---------------- ; split the fraction....
y'




(x-2)(y') (y+4)
5(x-2)^2 = ------- - ------- ; cancel top to bottom your y'
y' y'

(y+4)
5(x-2)^2 = (x-2) - ------- ; now subtract (x-2) from both sides
y'


- (y+4)
[5(x-2)^2]-(x-2) = ------- ; I forgot something on the paper....
y'

Answer 13

regardless tho, lets work this out:

-4-y
y' = ----------------
[5(x-2)^2] - (x-2)

try that one :)

Answer 14

(x=3,y=1)

y' = (-4-1) / [5(3-2)^2 - (3-2)]

y' = -5 / 4 , is that right?

Answer 15

HAH!!! I got the same answer as I did with the last one lol

Answer 16

yessss thank you so much