Ask your own question, for FREE!
Mathematics 17 Online
Parth (parthkohli):

What is implicit differentiation?

Parth (parthkohli):

@lgbasallote

Parth (parthkohli):

@Chlorophyll @sauravshakya @Hero

Parth (parthkohli):

@mukushla

Parth (parthkohli):

Sorry for tagging y'all, but I really need help.

OpenStudy (anonymous):

What do u want to know?

Parth (parthkohli):

I still am not getting it. Can you explain me what it is?

OpenStudy (anonymous):

Have u tried KHANACADEMY videos

OpenStudy (zzr0ck3r):

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)

OpenStudy (anonymous):

its great

OpenStudy (anonymous):

When y can't be expressed in terms of x "only". then y is implicit function of x

OpenStudy (zzr0ck3r):

its all about taking derivative without knowing explicitly what y depends on.

OpenStudy (zzr0ck3r):

for y = x y*(dy/dt) = 1 (dy/dt) = 1/y

Parth (parthkohli):

@cwrw238 do you have an idea?

OpenStudy (zzr0ck3r):

I swear im not lying.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Suppose \[ x^2 + 3 y^2 =1\] then 2 x + 6y y' =0 6yy'=-2 x y'=x/(3y)

Parth (parthkohli):

To be honest, I still don't get it. :(

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Read this http://tutorial.math.lamar.edu/Classes/CalcI/ImplicitDIff.aspx

OpenStudy (cwrw238):

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

Parth (parthkohli):

@eliassaab Yes, sir. The fact that I couldn't get the concept at Paul's Notes brought me here to ask this question.

OpenStudy (cwrw238):

you then find dy/dx in terms of x and y

OpenStudy (zzr0ck3r):

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)

OpenStudy (zzr0ck3r):

sorry dy/dx

OpenStudy (anonymous):

exactly, you only need to derive, what do you haven't understand, try to make an exercise, it helps!

OpenStudy (cwrw238):

example of implicit function is circle formula x^2 + y^2 = 9 which implies that y is a function of x

OpenStudy (anonymous):

try resolve that! :D

Parth (parthkohli):

Really sorry, guys. Haven't gotten it yet :(

OpenStudy (anonymous):

See the step 1) differentiate: 2x+2y*y' = 0 2) move y'=-x/y 3)have a party

OpenStudy (anonymous):

here you can see that it is really a circle! : http://www.wolframalpha.com/input/?i=y'=-x/y

Parth (parthkohli):

D':

OpenStudy (anonymous):

Step by step, slowly: y^2+x^3+x=1 what is the derivative?

Parth (parthkohli):

\[{dy^2 \over dx} + 3x^2 + 1 = 0\]

Parth (parthkohli):

I don't know if that is it... I'm just a beginner!

OpenStudy (anonymous):

Ok, I'm sorry, you're right, now derive the y

OpenStudy (anonymous):

dy^2/dx = ?

Parth (parthkohli):

I don't know...

OpenStudy (anonymous):

dy^2/dx = 2y*dy/dx using the chain rule, right?

Parth (parthkohli):

I don't understand! :'( Yes, I do know the chain rule... but I don't know the application of it here!!! ='(

Parth (parthkohli):

Where's the outside and inside function?

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

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

Parth (parthkohli):

@descripter37: Wait a second, I'm working on it on another site... I'd return back to you in a minute :)

OpenStudy (anonymous):

Ok, tell me if you get it :D

OpenStudy (anonymous):

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

Parth (parthkohli):

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 \(x\).

Parth (parthkohli):

I'm working on it... no need to waste your time more :P

Parth (parthkohli):

Thank you alL!

OpenStudy (anonymous):

Yuhuuuu sorry for all the time, but I'm not very well on English, I'm italian :3 good derivatives

Parth (parthkohli):

No worries. Ciao!

Parth (parthkohli):

Have you been to Venice?

Parth (parthkohli):

looool

OpenStudy (anonymous):

No, I'm close to Rome :)

Parth (parthkohli):

Is Milan close to Rome?

OpenStudy (anonymous):

Streets of italy ♥

Parth (parthkohli):

Considering that 580 kilometres is close :p

OpenStudy (anonymous):

Ehr, no it's very distant, I'm on the center of Italy but if 580 km is close... maybe yes!! XD

OpenStudy (anonymous):

Sigh, in all the world people think about Italy only north and south, mode, and pizza, us at the center are ignored! XD

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!