What is implicit differentiation?
@lgbasallote
@Chlorophyll @sauravshakya @Hero
@mukushla
Sorry for tagging y'all, but I really need help.
What do u want to know?
I still am not getting it. Can you explain me what it is?
Have u tried KHANACADEMY videos
um so if we implicit to y, we may not know what y(t) is. so we use chain rule and say y*(dy/dt)
its great
When y can't be expressed in terms of x "only". then y is implicit function of x
its all about taking derivative without knowing explicitly what y depends on.
for y = x y*(dy/dt) = 1 (dy/dt) = 1/y
@cwrw238 do you have an idea?
I swear im not lying.
what you do is differentiate the function: ex: y=x^2+1 dy/dx = 2x And it's done, but: y^2=x^2 d(y^2)/dx = 2x dy/dx * 2y = 2x so dy/dx = x/y
Suppose \[ x^2 + 3 y^2 =1\] then 2 x + 6y y' =0 6yy'=-2 x y'=x/(3y)
To be honest, I still don't get it. :(
you have to find the y' also called f'(x) or also dy/dx, so derivate by x and move thing to get your solution
an example is differentiate x^2 y + y^2 = 4 with respect to x you treat y as a function of x so y *2x + x^2*dy/dx + 2y * dy/dx = 0 using product rule and chain rule
@eliassaab Yes, sir. The fact that I couldn't get the concept at Paul's Notes brought me here to ask this question.
you then find dy/dx in terms of x and y
when we explicite, we are saying y = f(x) so y is dependent on x so when we explicite differentiate we are doing so with respect to the x that we know y depends on. now with implicite, we do not know that for y = x that y depends on x. so we make no assumtions when we differentiate. thus y = x y*(dy/dt) = 1 (lhs = chain rule)
sorry dy/dx
exactly, you only need to derive, what do you haven't understand, try to make an exercise, it helps!
example of implicit function is circle formula x^2 + y^2 = 9 which implies that y is a function of x
try resolve that! :D
Really sorry, guys. Haven't gotten it yet :(
See the step 1) differentiate: 2x+2y*y' = 0 2) move y'=-x/y 3)have a party
here you can see that it is really a circle! : http://www.wolframalpha.com/input/?i=y'=-x/y
D':
Step by step, slowly: y^2+x^3+x=1 what is the derivative?
\[{dy^2 \over dx} + 3x^2 + 1 = 0\]
I don't know if that is it... I'm just a beginner!
Ok, I'm sorry, you're right, now derive the y
dy^2/dx = ?
I don't know...
dy^2/dx = 2y*dy/dx using the chain rule, right?
I don't understand! :'( Yes, I do know the chain rule... but I don't know the application of it here!!! ='(
Where's the outside and inside function?
y is a function the derivative of y is the derivative like is a variable multiply the derivative of y, this is the chain rule right? (f(g(x)))'=f'(g(x))*g'(x)*x' you understand?
y is the outside function, the inside function is the part with the x, but at you doesn't interest, just writr y', you don't need to calculate it
@descripter37: Wait a second, I'm working on it on another site... I'd return back to you in a minute :)
Ok, tell me if you get it :D
Let f(x) = x^2 f^2(x) = x^4 .. Then, d/dx (f^2(x)) = 4x^3 ( As we know the basic rule ). from f(x), d/dx (f^2(x)) = d/dx (x^2)^2 = d/dx (f(x)^2) = 2 (f(x)) d/dx f(x) .. lol i hope you understand.. just clarified what was @decripter37 saying
Ok, so when we don't know what \(y \) is in respect to \(x\), we leave it like that and call it an unknown function of \(x\).
I'm working on it... no need to waste your time more :P
Thank you alL!
Yuhuuuu sorry for all the time, but I'm not very well on English, I'm italian :3 good derivatives
No worries. Ciao!
and remember http://fc05.deviantart.net/fs70/i/2011/088/f/6/don__t_drink_and_derive_by_fire_camel-d3cr4vs.png
Have you been to Venice?
looool
No, I'm close to Rome :)
Is Milan close to Rome?
Streets of italy ♥
Considering that 580 kilometres is close :p
Ehr, no it's very distant, I'm on the center of Italy but if 580 km is close... maybe yes!! XD
Sigh, in all the world people think about Italy only north and south, mode, and pizza, us at the center are ignored! XD
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