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remember product rule for implicit
\[(fg)' = f'g +fg'\]

M = slope = dy/dx

oh so for implicit we use the product?

always?

yes and chain rule....since it is with respect to x
\[d/dx f(y) = f'(y)*\frac{dy}{dx}\]

so which is f and which is g

im a little confused since xy^3 are together.

depends on the term...first there is "xy^3"
f(x) = x
g(y) = y^3

so f = x and g= y^3 ?

so is the derivative: x*3y+ y^3*1 ?

yes, so result is
\[1*y^{3}+ x*3y^{2} \frac{dy}{dx}\]

dy/dx?

thats what the slope is...which is what we are trying to find

all i know is the product rule is this:
\[\frac{ d }{ dx } (fg)= f g \prime+g f \prime\]

anytime you differentiate a function of "y" wrt x you must multiply by "dy/dx"
refer to my prev post

otay. so then what.

repeat process for all terms of equation

what do you mean

xy^3 +xy = 4
you have to differentiate each term, the "xy^3" and "xy" adn "4"

3xy^2.....

??
first term is
\[(y^{3} +3xy^{2} \frac{dy}{dx})\]

solve for "dy/dx" in terms of x and y

okay i understand from here:
\[\frac{ dy }{ dx}= 3xy ^{2}+ y ^{3}\]

then from there, i don't know what to do.

i didn't i'm doing this step by step

from the result.. ^^ i just multiplied.

?? sorry i am not understanding

you know.

hey mathcalculus

do you know chain rule

|dw:1363334172119:dw|

i get the feeling i lost you when i started using "dy/dx" :{

mhm :/

|dw:1363334256950:dw|

|dw:1363334523512:dw|

wait wait wait

remember how chain rule works

the chain rule: if f(x)= g(k(x)) then f prime x= g prime (k(x))k prime (x) correct?

yes that is chain rule

noob? -_-

411- not a guy.

haha trust me we are all noobs, just wait till you see how much those phd profs know

okay dc back to math.

thank god, no more crazy math after this. but seriously back to work lol

i have to understand this.

ok ok so for implicit differentiation when you see Y think about it like some function on x

ello?

yup got that.

then?

ok do this

|dw:1363334909090:dw|

yes you are going to need both those rules for this

no remember than y is like a function of x

OH

darn it.

|dw:1363334981033:dw|

so we use the product rule 1st.

sub 2x where you see Y and solve it

i have no idea what im doing.

for some reason I'm all confused.

okay look at this question again same question but im telling you what y is

don't tell me what y is. how am i suppose to know alone?

let's do that problem that I gave.

|dw:1363335163468:dw|

just do this im trying to teach you how to do the problem above and every other problem like this

k -.-

x= 2/5 ?

i plugged in 2x..

wut no find dervivative

chain rule

|dw:1363335349192:dw|

do you get how i got that when i differentiated there

whats confusing you there is it the d/dx?

i know that you know how to take derivate of Xs

you there?

any progress being made here...

yeah im looking.

ya i think so lol we've found the root problem here

she just doesnt know what it means to difference functions as opposed to variables

:)

|dw:1363335696884:dw|

do you understand this yet?

yeah...sorry i figured that was implied since we are dealing with implicit differentiation :|

thats chain rule remember you have to multiply by the derivative of the function inside the bracket

i understand you got the derivative

|dw:1363335825708:dw|

|dw:1363335821148:dw|

this is known as the chain rule

i remember the chain rule a little better now.

so if its |dw:1363335898535:dw|

ok lets go back to our problem knowing the chain rule now

gratzie

|dw:1363336027415:dw|

so for the implict up there

ok

got it, think the x as in y.

so our problem... is this: xy^3+xy=4 at the point (2,1) .

start by:

|dw:1363336084849:dw|

|dw:1363336247542:dw|

so you try it d/dx (xy^3) = ? if Y is a function of x