Question

use implicit integration to find dy/dx . y sin(1/y)= 1-xy

Answer 1

use the chain rule, and treat y as a differentiabe function of x.

Answer 2

i see chain rule, product rule, trig stuffs, ... yep

Answer 3

implicit is no different from explicit; derivative rules are the same regardless of what they want to name it

Answer 4

can you show me please?

Answer 5

show you the rules of derivatives?

Answer 6

no. how to solve it.

Answer 7

i spose i could ...

what the derivative of: xy ?

Answer 8

y i think. it said we must search dy/ dy by using implicit integration

Answer 9

its the product rule, can you show me the generic derivative of a product?

Answer 10

what is that. i'm sorry but i don;t know

Answer 11

if you dont know what the product rule is for derivatives, then you are asking the wrong question ....

Answer 12

i know what is product rule.

Answer 13

(fg)' = f'g+fg'

Answer 14

notations differ which is why i want you to write up what you know it as

Answer 15

\[y \sin (1/y) = 1-xy\] find thedy/dx by using the implicit derivatives.

Answer 16

yes, thats what you wrote to begin with

and at the moment we need to work it out, piece by piece so that you get a better concept of it

Answer 17

lets start with the simplest part of it and derive: xy

Answer 18

x dy/dx+ y. right??

Answer 19

yes

Answer 20

and the derivative of 1 is always 0

so that takes care of the right side.

y sin(1/y) = xy' + y

the left side is a product rule as well

Answer 21

i know the concept already. but to derive ysin (1/y) i stuck already

Answer 22

forget the we are looking at a sin(...) and product it to begin with

Answer 23

[y sin(1/y)]' = y' sin(1/y) + y sin'(1/y)

thats simple enough; now we can focus on what sin'(1/y) needs to be

Answer 24

you agree with that so far?

Answer 25

yes yes. but then when to put dy/dx afer derive the 1/y?

Answer 26

when? the same time you do it normally :)

so far we have this

\[\frac{dy}{dx}sin(\frac{1}{y})+y\frac{d}{dx}(sin(\frac{1}{y}))=x\frac{dy}{dx}+y\]
the final is a chain rule
\[\frac{d}{dx}(sin(\frac{1}{y}))\to\ cos(\frac{1}{y})\ \frac{d}{dx}(\frac{1}{y})\]
\[\frac{d}{dx}(\frac{1}{y})\to\ \frac{dy}{dx}(\frac{-1}{y^2})\]

Answer 27

apparently, someone thought it a good idea to have the switch for the wireless internet to be placed on the keypad itself so that when you hit it by accident; you get disconnected :/

Answer 28

what do you mean?

Answer 29

i mean what i said. my wireless internet switch is in a bad spot.

Answer 30

mine too. never mind.

Answer 31

:)

all those dy/dx stuff crowd the scene, i like the newton notation better for this:

\[y'\ sin(\frac{1}{y}) + -(\frac{y}{y^2}) cos(\frac{1}{y})\ y' = xy' + y\]

the rest is just algebra to isolate the y' s

Answer 32

ok. got it. thank you. :)

Answer 33

youre welcome; and good luck