Question

Help:implicit differentiation with a fractions :(

Answer 1

multiplying everything by 36 you get;
2x^2+y^2=36
which maybe more clear for you, :)

Answer 2

hey! :) i tried this: x²/16 + y²/36 = 1
(1/16)x² + (1/36)y² = 1
2(1/16)x + 2(1/36)yy' = 0

Answer 3

but so far im not sure what to do after because i know i get x/8 + ?? but then i can't really multiply (1/36) 2y*dy/dx ?

Answer 4

isn't the point here to find dy/dx?

Answer 5

let \(F(x,y)=2x^2+y^2=36\)
then:
\(dF(x,y)=4xdx+2ydy=0\)
from here:
\(y'=dy/dx=-4x/2y=-2x/y\)

Answer 6

huh? o.o

Answer 7

howd you get this??
let F(x,y)=2x2+y2=36

Answer 8

\(dF(x,y)\) is colled full differential
it is: \(dF(x,y)=F_xdx+F_ydy\)

Answer 9

i never seen that before

Answer 10

\(F_x\) is partial derivative respect to x

Answer 11

ok

Answer 12

i know those, but where are the numbers coming from?

Answer 13

in what order, where did you take them?

Answer 14

i know they're given but how did you get 36 on the other side. and 2x^2+y^2let F(x,y)=2x2+y2=36

Answer 15

ok, other way:
think that y is actualy a function of x, so it is y=y(x). Then:
\(2x^2+y(x)^2=36\)
now we differentiate keeping this in mind
4x+2yy'=0
now just solve for y'

Answer 16

2x2+y(x)2=36?

Answer 17

yes i know when to differentiate but how did you change the function to that

Answer 18

Ya you right I made a mistake. I was treating 36 like 32, :). Shold be. Rest of steps same.
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