anonymous
  • anonymous
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:
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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dumbcow
  • dumbcow
remember product rule for implicit \[(fg)' = f'g +fg'\]
dumbcow
  • dumbcow
M = slope = dy/dx
anonymous
  • anonymous
oh so for implicit we use the product?

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anonymous
  • anonymous
always?
dumbcow
  • dumbcow
yes and chain rule....since it is with respect to x \[d/dx f(y) = f'(y)*\frac{dy}{dx}\]
anonymous
  • anonymous
so which is f and which is g
anonymous
  • anonymous
im a little confused since xy^3 are together.
dumbcow
  • dumbcow
depends on the term...first there is "xy^3" f(x) = x g(y) = y^3
anonymous
  • anonymous
so f = x and g= y^3 ?
anonymous
  • anonymous
so is the derivative: x*3y+ y^3*1 ?
dumbcow
  • dumbcow
yes, so result is \[1*y^{3}+ x*3y^{2} \frac{dy}{dx}\]
anonymous
  • anonymous
dy/dx?
dumbcow
  • dumbcow
thats what the slope is...which is what we are trying to find
anonymous
  • anonymous
all i know is the product rule is this: \[\frac{ d }{ dx } (fg)= f g \prime+g f \prime\]
dumbcow
  • dumbcow
anytime you differentiate a function of "y" wrt x you must multiply by "dy/dx" refer to my prev post
anonymous
  • anonymous
otay. so then what.
dumbcow
  • dumbcow
repeat process for all terms of equation
anonymous
  • anonymous
?
anonymous
  • anonymous
what do you mean
dumbcow
  • dumbcow
xy^3 +xy = 4 you have to differentiate each term, the "xy^3" and "xy" adn "4"
anonymous
  • anonymous
3xy^2.....
dumbcow
  • dumbcow
?? first term is \[(y^{3} +3xy^{2} \frac{dy}{dx})\]
dumbcow
  • 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
  • dumbcow
solve for "dy/dx" in terms of x and y
anonymous
  • anonymous
okay i understand from here: \[\frac{ dy }{ dx}= 3xy ^{2}+ y ^{3}\]
anonymous
  • anonymous
then from there, i don't know what to do.
anonymous
  • anonymous
@dumbcow
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
  • anonymous
i didn't i'm doing this step by step
anonymous
  • anonymous
from the result.. ^^ i just multiplied.
dumbcow
  • dumbcow
?? sorry i am not understanding
anonymous
  • anonymous
remember before you wrote "yes, result is bla blah" then from there i just multiplied... x*(3y)^2 + y^3 *1
anonymous
  • anonymous
you know.
dan815
  • dan815
hey mathcalculus
dan815
  • dan815
do you know chain rule
dan815
  • dan815
|dw:1363334172119:dw|
dumbcow
  • 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
  • dumbcow
i get the feeling i lost you when i started using "dy/dx" :{
anonymous
  • anonymous
mhm :/
anonymous
  • anonymous
|dw:1363334405957:dw| @dan815
dan815
  • dan815
|dw:1363334256950:dw|
dan815
  • 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
  • dan815
|dw:1363334523512:dw|
anonymous
  • anonymous
wait wait wait
dumbcow
  • dumbcow
thx @dan815 for clarifying :) i didn't explain very well
dan815
  • dan815
remember how chain rule works
anonymous
  • anonymous
the chain rule: if f(x)= g(k(x)) then f prime x= g prime (k(x))k prime (x) correct?
anonymous
  • anonymous
@dumbcow you explained very good too.
dan815
  • dan815
@dumbcow no its okay lol i can relate to him cuz i know how noobs think :)
anonymous
  • anonymous
i just can't understand @dan815 techwriting lol
dan815
  • dan815
yes that is chain rule
anonymous
  • anonymous
noob? -_-
anonymous
  • anonymous
411- not a guy.
dan815
  • dan815
haha trust me we are all noobs, just wait till you see how much those phd profs know
anonymous
  • anonymous
okay dc back to math.
anonymous
  • anonymous
thank god, no more crazy math after this. but seriously back to work lol
anonymous
  • anonymous
i have to understand this.
dan815
  • dan815
ok ok so for implicit differentiation when you see Y think about it like some function on x
anonymous
  • anonymous
ello?
anonymous
  • anonymous
yup got that.
anonymous
  • anonymous
then?
dan815
  • dan815
ok do this
anonymous
  • 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
  • dan815
|dw:1363334909090:dw|
dan815
  • dan815
yes you are going to need both those rules for this
dan815
  • dan815
no remember than y is like a function of x
anonymous
  • anonymous
OH
anonymous
  • anonymous
darn it.
dan815
  • dan815
|dw:1363334981033:dw|
anonymous
  • anonymous
so we use the product rule 1st.
dan815
  • dan815
sub 2x where you see Y and solve it
anonymous
  • anonymous
i have no idea what im doing.
anonymous
  • anonymous
for some reason I'm all confused.
dan815
  • dan815
okay look at this question again same question but im telling you what y is
anonymous
  • anonymous
don't tell me what y is. how am i suppose to know alone?
anonymous
  • anonymous
let's do that problem that I gave.
dan815
  • dan815
|dw:1363335163468:dw|
dan815
  • dan815
just do this im trying to teach you how to do the problem above and every other problem like this
anonymous
  • anonymous
k -.-
anonymous
  • anonymous
x= 2/5 ?
anonymous
  • anonymous
i plugged in 2x..
dan815
  • dan815
wut no find dervivative
dan815
  • dan815
chain rule
anonymous
  • anonymous
?
dan815
  • dan815
|dw:1363335349192:dw|
dan815
  • dan815
do you get how i got that when i differentiated there
dan815
  • dan815
whats confusing you there is it the d/dx?
dan815
  • dan815
i know that you know how to take derivate of Xs
dan815
  • dan815
you there?
dumbcow
  • dumbcow
any progress being made here...
anonymous
  • anonymous
yeah im looking.
dan815
  • dan815
ya i think so lol we've found the root problem here
dan815
  • dan815
she just doesnt know what it means to difference functions as opposed to variables
anonymous
  • anonymous
@dumbcow you used the product rule first which i was understanding.... until mr. CHAIN RULE came along to make a knot.
anonymous
  • anonymous
:)
dan815
  • dan815
|dw:1363335696884:dw|
dan815
  • dan815
do you understand this yet?
dumbcow
  • dumbcow
yeah...sorry i figured that was implied since we are dealing with implicit differentiation :|
anonymous
  • anonymous
yeah @dan815 but now why did you multiply 2 at the end.?
dan815
  • dan815
thats chain rule remember you have to multiply by the derivative of the function inside the bracket
anonymous
  • anonymous
i understand you got the derivative
anonymous
  • anonymous
|dw:1363335825708:dw|
dan815
  • dan815
|dw:1363335821148:dw|
dan815
  • dan815
this is known as the chain rule
anonymous
  • anonymous
okie let's do our problem @dan815
anonymous
  • 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
  • anonymous
i remember the chain rule a little better now.
dan815
  • dan815
so if its |dw:1363335898535:dw|
dan815
  • dan815
ok lets go back to our problem knowing the chain rule now
anonymous
  • anonymous
gratzie
dan815
  • 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
  • dan815
|dw:1363336027415:dw|
dan815
  • dan815
so for the implict up there
anonymous
  • anonymous
ok
anonymous
  • anonymous
got it, think the x as in y.
anonymous
  • anonymous
so our problem... is this: xy^3+xy=4 at the point (2,1) .
anonymous
  • anonymous
start by:
dan815
  • dan815
|dw:1363336084849:dw|
dan815
  • dan815
|dw:1363336247542:dw|
dan815
  • dan815
so you try it d/dx (xy^3) = ? if Y is a function of x
dan815
  • 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
  • anonymous
|dw:1363336482888:dw|
anonymous
  • anonymous
|dw:1363336550835:dw|
dan815
  • dan815
but Y isnt just a variable like X its a function so u have to use chain rule
dan815
  • dan815
ok please try this example
anonymous
  • anonymous
fine
dan815
  • dan815
|dw:1363336642423:dw|
anonymous
  • anonymous
what does that say....... ''/
anonymous
  • anonymous
is that a 2, an h, x ?
dan815
  • dan815
|dw:1363336742987:dw|