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mathcalculus
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:
remember product rule for implicit \[(fg)' = f'g +fg'\]
oh so for implicit we use the product?
yes and chain rule....since it is with respect to x \[d/dx f(y) = f'(y)*\frac{dy}{dx}\]
so which is f and which is g
im a little confused since xy^3 are together.
depends on the term...first there is "xy^3" f(x) = x g(y) = y^3
so f = x and g= y^3 ?
so is the derivative: x*3y+ y^3*1 ?
yes, so result is \[1*y^{3}+ x*3y^{2} \frac{dy}{dx}\]
thats what the slope is...which is what we are trying to find
all i know is the product rule is this: \[\frac{ d }{ dx } (fg)= f g \prime+g f \prime\]
anytime you differentiate a function of "y" wrt x you must multiply by "dy/dx" refer to my prev post
otay. so then what.
repeat process for all terms of equation
xy^3 +xy = 4 you have to differentiate each term, the "xy^3" and "xy" adn "4"
?? first term is \[(y^{3} +3xy^{2} \frac{dy}{dx})\]
next term is "xy" .... derivative is \[(y + x \frac{dy}{dx})\] then of course , derivative of constant is 0 now you have \[y^{3} +3xy^{2} \frac{dy}{dx} +y+x \frac{dy}{dx} = 0\]
solve for "dy/dx" in terms of x and y
okay i understand from here: \[\frac{ dy }{ dx}= 3xy ^{2}+ y ^{3}\]
then from there, i don't know what to do.
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
i didn't i'm doing this step by step
from the result.. ^^ i just multiplied.
?? sorry i am not understanding
remember before you wrote "yes, result is bla blah" then from there i just multiplied... x*(3y)^2 + y^3 *1
let me just show the steps, maybe that will help \[y^{3}+3xy^{2} \frac{dy}{dx} +y+x \frac{dy}{dx} = 0\] \[3xy^{2} \frac{dy}{dx}+x \frac{dy}{dx} = -y^{3}-y\] \[\frac{dy}{dx}(3xy^{2}+x) = -y^{3}-y\] \[\frac{dy}{dx} = \frac{-y^{3}-y}{3xy^{2}+x}\]
i get the feeling i lost you when i started using "dy/dx" :{
|dw:1363334405957: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?
thx @dan815 for clarifying :) i didn't explain very well
remember how chain rule works
the chain rule: if f(x)= g(k(x)) then f prime x= g prime (k(x))k prime (x) correct?
@dumbcow you explained very good too.
@dumbcow no its okay lol i can relate to him cuz i know how noobs think :)
i just can't understand @dan815 techwriting lol
haha trust me we are all noobs, just wait till you see how much those phd profs know
okay dc back to math.
thank god, no more crazy math after this. but seriously back to work lol
i have to understand this.
ok ok so for implicit differentiation when you see Y think about it like some function on x
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.
yes you are going to need both those rules for this
no remember than y is like a function of x
so we use the product rule 1st.
sub 2x where you see Y and solve it
i have no idea what im doing.
for some reason I'm all confused.
okay look at this question again same question but im telling you what y is
don't tell me what y is. how am i suppose to know alone?
let's do that problem that I gave.
just do this im trying to teach you how to do the problem above and every other problem like this
do you get how i got that when i differentiated there
whats confusing you there is it the d/dx?
i know that you know how to take derivate of Xs
any progress being made here...
ya i think so lol we've found the root problem here
she just doesnt know what it means to difference functions as opposed to variables
@dumbcow you used the product rule first which i was understanding.... until mr. CHAIN RULE came along to make a knot.
do you understand this yet?
yeah...sorry i figured that was implied since we are dealing with implicit differentiation :|
yeah @dan815 but now why did you multiply 2 at the end.?
thats chain rule remember you have to multiply by the derivative of the function inside the bracket
i understand you got the derivative
|dw:1363335825708:dw|
this is known as the chain rule
okie let's do our problem @dan815
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.
i remember the chain rule a little better now.
so if its |dw:1363335898535:dw|
ok lets go back to our problem knowing the chain rule now
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
so for the implict up there
got it, think the x as in y.
so our problem... is this: xy^3+xy=4 at the point (2,1) .