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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:
 one year ago
 one year ago
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:
 one year ago
 one year ago

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dumbcowBest ResponseYou've already chosen the best response.4
remember product rule for implicit \[(fg)' = f'g +fg'\]
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
oh so for implicit we use the product?
 one year ago

dumbcowBest ResponseYou've already chosen the best response.4
yes and chain rule....since it is with respect to x \[d/dx f(y) = f'(y)*\frac{dy}{dx}\]
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
so which is f and which is g
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
im a little confused since xy^3 are together.
 one year ago

dumbcowBest ResponseYou've already chosen the best response.4
depends on the term...first there is "xy^3" f(x) = x g(y) = y^3
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
so f = x and g= y^3 ?
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
so is the derivative: x*3y+ y^3*1 ?
 one year ago

dumbcowBest ResponseYou've already chosen the best response.4
yes, so result is \[1*y^{3}+ x*3y^{2} \frac{dy}{dx}\]
 one year ago

dumbcowBest ResponseYou've already chosen the best response.4
thats what the slope is...which is what we are trying to find
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
all i know is the product rule is this: \[\frac{ d }{ dx } (fg)= f g \prime+g f \prime\]
 one year ago

dumbcowBest ResponseYou've already chosen the best response.4
anytime you differentiate a function of "y" wrt x you must multiply by "dy/dx" refer to my prev post
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
otay. so then what.
 one year ago

dumbcowBest ResponseYou've already chosen the best response.4
repeat process for all terms of equation
 one year ago

dumbcowBest ResponseYou've already chosen the best response.4
xy^3 +xy = 4 you have to differentiate each term, the "xy^3" and "xy" adn "4"
 one year ago

dumbcowBest ResponseYou've already chosen the best response.4
?? first term is \[(y^{3} +3xy^{2} \frac{dy}{dx})\]
 one year ago

dumbcowBest ResponseYou've already chosen the best response.4
next term is "xy" .... derivative is \[(y + x \frac{dy}{dx})\] then of course , derivative of constant is 0 now you have \[y^{3} +3xy^{2} \frac{dy}{dx} +y+x \frac{dy}{dx} = 0\]
 one year ago

dumbcowBest ResponseYou've already chosen the best response.4
solve for "dy/dx" in terms of x and y
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
okay i understand from here: \[\frac{ dy }{ dx}= 3xy ^{2}+ y ^{3}\]
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
then from there, i don't know what to do.
 one year ago

dumbcowBest ResponseYou've already chosen the best response.4
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
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
i didn't i'm doing this step by step
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
from the result.. ^^ i just multiplied.
 one year ago

dumbcowBest ResponseYou've already chosen the best response.4
?? sorry i am not understanding
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
remember before you wrote "yes, result is bla blah" then from there i just multiplied... x*(3y)^2 + y^3 *1
 one year ago

dumbcowBest ResponseYou've already chosen the best response.4
let me just show the steps, maybe that will help \[y^{3}+3xy^{2} \frac{dy}{dx} +y+x \frac{dy}{dx} = 0\] \[3xy^{2} \frac{dy}{dx}+x \frac{dy}{dx} = y^{3}y\] \[\frac{dy}{dx}(3xy^{2}+x) = y^{3}y\] \[\frac{dy}{dx} = \frac{y^{3}y}{3xy^{2}+x}\]
 one year ago

dumbcowBest ResponseYou've already chosen the best response.4
i get the feeling i lost you when i started using "dy/dx" :{
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
dw:1363334405957:dw @dan815
 one year ago

dan815Best ResponseYou've already chosen the best response.1
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?
 one year ago

dumbcowBest ResponseYou've already chosen the best response.4
thx @dan815 for clarifying :) i didn't explain very well
 one year ago

dan815Best ResponseYou've already chosen the best response.1
remember how chain rule works
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
the chain rule: if f(x)= g(k(x)) then f prime x= g prime (k(x))k prime (x) correct?
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
@dumbcow you explained very good too.
 one year ago

dan815Best ResponseYou've already chosen the best response.1
@dumbcow no its okay lol i can relate to him cuz i know how noobs think :)
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
i just can't understand @dan815 techwriting lol
 one year ago

dan815Best ResponseYou've already chosen the best response.1
haha trust me we are all noobs, just wait till you see how much those phd profs know
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
okay dc back to math.
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
thank god, no more crazy math after this. but seriously back to work lol
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
i have to understand this.
 one year ago

dan815Best ResponseYou've already chosen the best response.1
ok ok so for implicit differentiation when you see Y think about it like some function on x
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
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.
 one year ago

dan815Best ResponseYou've already chosen the best response.1
yes you are going to need both those rules for this
 one year ago

dan815Best ResponseYou've already chosen the best response.1
no remember than y is like a function of x
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
so we use the product rule 1st.
 one year ago

dan815Best ResponseYou've already chosen the best response.1
sub 2x where you see Y and solve it
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
i have no idea what im doing.
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
for some reason I'm all confused.
 one year ago

dan815Best ResponseYou've already chosen the best response.1
okay look at this question again same question but im telling you what y is
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
don't tell me what y is. how am i suppose to know alone?
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
let's do that problem that I gave.
 one year ago

dan815Best ResponseYou've already chosen the best response.1
just do this im trying to teach you how to do the problem above and every other problem like this
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
i plugged in 2x..
 one year ago

dan815Best ResponseYou've already chosen the best response.1
wut no find dervivative
 one year ago

dan815Best ResponseYou've already chosen the best response.1
do you get how i got that when i differentiated there
 one year ago

dan815Best ResponseYou've already chosen the best response.1
whats confusing you there is it the d/dx?
 one year ago

dan815Best ResponseYou've already chosen the best response.1
i know that you know how to take derivate of Xs
 one year ago

dumbcowBest ResponseYou've already chosen the best response.4
any progress being made here...
 one year ago

dan815Best ResponseYou've already chosen the best response.1
ya i think so lol we've found the root problem here
 one year ago

dan815Best ResponseYou've already chosen the best response.1
she just doesnt know what it means to difference functions as opposed to variables
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
@dumbcow you used the product rule first which i was understanding.... until mr. CHAIN RULE came along to make a knot.
 one year ago

dan815Best ResponseYou've already chosen the best response.1
do you understand this yet?
 one year ago

dumbcowBest ResponseYou've already chosen the best response.4
yeah...sorry i figured that was implied since we are dealing with implicit differentiation :
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
yeah @dan815 but now why did you multiply 2 at the end.?
 one year ago

dan815Best ResponseYou've already chosen the best response.1
thats chain rule remember you have to multiply by the derivative of the function inside the bracket
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
i understand you got the derivative
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
dw:1363335825708:dw
 one year ago

dan815Best ResponseYou've already chosen the best response.1
this is known as the chain rule
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
okie let's do our problem @dan815
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
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.
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
i remember the chain rule a little better now.
 one year ago

dan815Best ResponseYou've already chosen the best response.1
so if its dw:1363335898535:dw
 one year ago

dan815Best ResponseYou've already chosen the best response.1
ok lets go back to our problem knowing the chain rule now
 one year ago

dan815Best ResponseYou've already chosen the best response.1
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
 one year ago

dan815Best ResponseYou've already chosen the best response.1
so for the implict up there
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
got it, think the x as in y.
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
so our problem... is this: xy^3+xy=4 at the point (2,1) .
 one year ago