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 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
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:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.4well 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

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

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

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

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

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

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

dan815
 one year ago
Best ResponseYou've already chosen the best response.1think 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?

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

dan815
 one year ago
Best ResponseYou've already chosen the best response.1remember how chain rule works

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

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

dan815
 one year ago
Best ResponseYou've already chosen the best response.1@dumbcow no its okay lol i can relate to him cuz i know how noobs think :)

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

dan815
 one year ago
Best ResponseYou've already chosen the best response.1haha trust me we are all noobs, just wait till you see how much those phd profs know

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

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

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

dan815
 one year ago
Best ResponseYou've already chosen the best response.1ok ok so for implicit differentiation when you see Y think about it like some function on x

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.1yes you are going to need both those rules for this

dan815
 one year ago
Best ResponseYou've already chosen the best response.1no remember than y is like a function of x

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

dan815
 one year ago
Best ResponseYou've already chosen the best response.1sub 2x where you see Y and solve it

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

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

dan815
 one year ago
Best ResponseYou've already chosen the best response.1okay look at this question again same question but im telling you what y is

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

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

dan815
 one year ago
Best ResponseYou've already chosen the best response.1just do this im trying to teach you how to do the problem above and every other problem like this

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

dan815
 one year ago
Best ResponseYou've already chosen the best response.1wut no find dervivative

dan815
 one year ago
Best ResponseYou've already chosen the best response.1do you get how i got that when i differentiated there

dan815
 one year ago
Best ResponseYou've already chosen the best response.1whats confusing you there is it the d/dx?

dan815
 one year ago
Best ResponseYou've already chosen the best response.1i know that you know how to take derivate of Xs

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

dan815
 one year ago
Best ResponseYou've already chosen the best response.1ya i think so lol we've found the root problem here

dan815
 one year ago
Best ResponseYou've already chosen the best response.1she just doesnt know what it means to difference functions as opposed to variables

mathcalculus
 one year ago
Best 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.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1do you understand this yet?

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

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

dan815
 one year ago
Best ResponseYou've already chosen the best response.1thats chain rule remember you have to multiply by the derivative of the function inside the bracket

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

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

dan815
 one year ago
Best ResponseYou've already chosen the best response.1this is known as the chain rule

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

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i 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.

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

dan815
 one year ago
Best ResponseYou've already chosen the best response.1so if its dw:1363335898535:dw

dan815
 one year ago
Best ResponseYou've already chosen the best response.1ok lets go back to our problem knowing the chain rule now

dan815
 one year ago
Best ResponseYou've already chosen the best response.1what 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
 one year ago
Best ResponseYou've already chosen the best response.1so for the implict up there

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

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