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mathcalculus
 2 years 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:
mathcalculus
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.4remember product rule for implicit \[(fg)' = f'g +fg'\]

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

dumbcow
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0so which is f and which is g

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

dumbcow
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0so f = x and g= y^3 ?

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

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

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

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

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

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

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

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

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

dumbcow
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0i didn't i'm doing this step by step

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

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

mathcalculus
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.4i get the feeling i lost you when i started using "dy/dx" :{

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

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

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

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

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

dan815
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0okay dc back to math.

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

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

dan815
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1yes you are going to need both those rules for this

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

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

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

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

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

dan815
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0don't tell me what y is. how am i suppose to know alone?

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

dan815
 2 years 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

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

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

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

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

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

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

mathcalculus
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1do you understand this yet?

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

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

dan815
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0i understand you got the derivative

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

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

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

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

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

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

dan815
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1so for the implict up there

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

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