HELP Please. Use implicit differentiation to find the equation of the tangent line to the curve xy^3+xy=4 at the point (2,1) . The equation of this tangent line can be written in the form: y=mx+b
where M is:
and where B is:

- anonymous

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- dumbcow

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

- dumbcow

M = slope = dy/dx

- anonymous

oh so for implicit we use the product?

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## More answers

- anonymous

always?

- dumbcow

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

- anonymous

so which is f and which is g

- anonymous

im a little confused since xy^3 are together.

- dumbcow

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

- anonymous

so f = x and g= y^3 ?

- anonymous

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

- dumbcow

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

- anonymous

dy/dx?

- dumbcow

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

- anonymous

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

- dumbcow

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

- anonymous

otay. so then what.

- dumbcow

repeat process for all terms of equation

- anonymous

?

- anonymous

what do you mean

- dumbcow

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

- anonymous

3xy^2.....

- dumbcow

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

- dumbcow

next term is "xy" .... derivative is
\[(y + x \frac{dy}{dx})\]
then of course , derivative of constant is 0
now you have
\[y^{3} +3xy^{2} \frac{dy}{dx} +y+x \frac{dy}{dx} = 0\]

- dumbcow

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

- anonymous

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

- anonymous

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

- anonymous

@dumbcow

- dumbcow

well from there you would plug in the given point (2,1) to give you the slope (M) of tangent line
but how did you solve for dy/dx .... your answer is wrong

- anonymous

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

- anonymous

from the result.. ^^ i just multiplied.

- dumbcow

?? sorry i am not understanding

- anonymous

remember before you wrote "yes, result is bla blah" then from there i just multiplied... x*(3y)^2 + y^3 *1

- anonymous

you know.

- dan815

hey mathcalculus

- dan815

do you know chain rule

- dan815

|dw:1363334172119:dw|

- dumbcow

let me just show the steps, maybe that will help
\[y^{3}+3xy^{2} \frac{dy}{dx} +y+x \frac{dy}{dx} = 0\]
\[3xy^{2} \frac{dy}{dx}+x \frac{dy}{dx} = -y^{3}-y\]
\[\frac{dy}{dx}(3xy^{2}+x) = -y^{3}-y\]
\[\frac{dy}{dx} = \frac{-y^{3}-y}{3xy^{2}+x}\]

- dumbcow

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

- anonymous

mhm :/

- anonymous

|dw:1363334405957:dw| @dan815

- dan815

|dw:1363334256950:dw|

- dan815

think about it like this Y is a function of X right so
say you have to take the derivative of 2x^2 what is it?

- dan815

|dw:1363334523512:dw|

- anonymous

wait wait wait

- dumbcow

thx @dan815 for clarifying :)
i didn't explain very well

- dan815

remember how chain rule works

- anonymous

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

- anonymous

@dumbcow you explained very good too.

- dan815

@dumbcow no its okay lol i can relate to him cuz i know how noobs think :)

- anonymous

i just can't understand @dan815 techwriting lol

- dan815

yes that is chain rule

- anonymous

noob? -_-

- anonymous

411- not a guy.

- dan815

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

- anonymous

okay dc back to math.

- anonymous

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

- anonymous

i have to understand this.

- dan815

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

- anonymous

ello?

- anonymous

yup got that.

- anonymous

then?

- dan815

ok do this

- anonymous

so first i was told to use the product rule.. then the chain. which that confused me a little. so lets take it step by step please.

- dan815

|dw:1363334909090:dw|

- dan815

yes you are going to need both those rules for this

- dan815

no remember than y is like a function of x

- anonymous

OH

- anonymous

darn it.

- dan815

|dw:1363334981033:dw|

- anonymous

so we use the product rule 1st.

- dan815

sub 2x where you see Y and solve it

- anonymous

i have no idea what im doing.

- anonymous

for some reason I'm all confused.

- dan815

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

- anonymous

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

- anonymous

let's do that problem that I gave.

- dan815

|dw:1363335163468:dw|

- dan815

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

- anonymous

k -.-

- anonymous

x= 2/5 ?

- anonymous

i plugged in 2x..

- dan815

wut no find dervivative

- dan815

chain rule

- anonymous

?

- dan815

|dw:1363335349192:dw|

- dan815

do you get how i got that when i differentiated there

- dan815

whats confusing you there is it the d/dx?

- dan815

i know that you know how to take derivate of Xs

- dan815

you there?

- dumbcow

any progress being made here...

- anonymous

yeah im looking.

- dan815

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

- dan815

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

- anonymous

@dumbcow you used the product rule first which i was understanding.... until mr. CHAIN RULE came along to make a knot.

- anonymous

:)

- dan815

|dw:1363335696884:dw|

- dan815

do you understand this yet?

- dumbcow

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

- anonymous

yeah @dan815 but now why did you multiply 2 at the end.?

- dan815

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

- anonymous

i understand you got the derivative

- anonymous

|dw:1363335825708:dw|

- dan815

|dw:1363335821148:dw|

- dan815

this is known as the chain rule

- anonymous

okie let's do our problem @dan815

- anonymous

i know youre trying to teach me, so i can do the rest.. etc; but i need to do one problem first in order to proceed.

- anonymous

i remember the chain rule a little better now.

- dan815

so if its |dw:1363335898535:dw|

- dan815

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

- anonymous

gratzie

- dan815

what i really wanted you to undetsand was this concept though basically you have to think of Y as a function so when you see Ys differentiating Y is like differentiation of a function so think chain rules

- dan815

|dw:1363336027415:dw|

- dan815

so for the implict up there

- anonymous

ok

- anonymous

got it, think the x as in y.

- anonymous

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

- anonymous

start by:

- dan815

|dw:1363336084849:dw|

- dan815

|dw:1363336247542:dw|

- dan815

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

- dan815

|dw:1363336439304:dw| there i put it brackets to emphasize thats Y really in your head you should imagine some functions with like X^4 + 3x+x....

- anonymous

|dw:1363336482888:dw|

- anonymous

|dw:1363336550835:dw|

- dan815

but Y isnt just a variable like X its a function so u have to use chain rule

- dan815

ok please try this example

- anonymous

fine

- dan815

|dw:1363336642423:dw|

- anonymous

what does that say....... ''/

- anonymous

is that a 2, an h, x ?

- dan815

|dw:1363336742987:dw|