CALCULUS 181 (IMPLICIT DIFFERENTIATION) WITH RESPECT X

[CHAMI]

Find dy/dx by implicit differentiation

x to the third power minus 3x to the 2nd power y + 2x y to the 2nd power = 12

x ^ 3 - 3x^2y + 2xy^2 = 12


We have xy terms that is what makes it more confuse because book does not show an example of that kind! I am having hard time solving this. Basically do not have a clue in other words! What is in the book does not make no sense at all. :roll: :roll: :roll: :roll: :roll:

Anywas, if I do it I would do like this which is this much I got idea after reading the book!

x^3 - 3x^2y + 2xy^2 = 12

After getting the derivitive of all

dy
__ [ 3x^2 - 6xy + 4xy ] = 0
dx

dy
___ [-6xy + 4xy] = -3x^2
dx

dy
___ [-6xy + 4xy]
dx ___________ = (-3x^2) / (-6xy +4xy)
[-6xy + 4xy]

This is wrong but I am glad anyone can help out! This much I have clue! :roll:

 

[tkhunny]

Your notation needs serious upgrading. After you introduced dy/dx, I hav no idea what you mean.

x^3 - 3x^2y + 2xy^2 = 12

One piece at a time.

Power Rule is sufficient
x^3 ==> 3*x^2

You need the Product Rule AND the Chain Rule on this one.
- 3x^2y ==> -2*x^2*(dy/dx) - 6*x*y

It is assumed that y=f(x), some function of 'x'. We just don't know what function.
2xy^2 ==> 2*x*(2*y*(dy/dx)) + 2*y^2

Put it all together and solve for dy/dx.

 

[Gene]

Nitpicking.
I would say proper notation would require
d(x^3) = 3x^2*dx
d(-3x^2y) ==> -3*x^2*dy - 6*x*y*dx
The shortcut division by dx is... distastefull.
---------------
Gene

 

[CHAMI]

Nitpicking.
I would say proper notation would require
d(x^3) = 3x^2*dx
d(-3x^2y) ==> -3*x^2*dy - 6*x*y*dx
The shortcut division by dx is... distastefull.
---------------
Gene

:roll: I am still confused both of your postings this part though

"

You need the Product Rule AND the Chain Rule on this one.
- 3x^2y ==> -2*x^2*(dy/dx) - 6*x*y " :shock:

Okay so what are we doing here is get the derivitive of whole equation and set it = 0. Then how do we figure out DY/DX like you use only power rule to get the derivitive of 3x^2 :roll: and use dy/dx for other ones, why?


Thank god if you can show the whole process

I will half and let me know too it is correct or not!

x^3 - 3x^2y + 2xy^2 = 12

3x^2 - 6xy dy/dx + 2x(2y) dy/dx = 0

3x^2 - dy/dx { 6xy + 4xy} = 0

-dy/dx {6xy+4xy} = -3x^2
divide both sides by -6xy+4xy gives:::::

dy/dx = -3x^2/-6xy+4xy //

WHAT IS THE DIFFERENCE BETWEEN D/DX AND DY/DX DO WE USE DY/DX WHEN WE HAVE XY TERMS TOGETHER? WHEN WE HAVE X TERM, WE USE D/DX ONLY? IS THAT HOW IT WORKS?

Is this correct? :shock: :shock: :shock: :shock: :shock: I am not a professional MATH champion

 

[Gene]

The product rule is necessary whenever you have two variables (x&y).
d(xy) = x*dy+y*dx
or your
d((-3x^2)(y)) ==> (-3*x^2)*(dy) + ((-6*x)*(y))*dx
TKH knew you were heading for dy/dx so he did the dx division
-3*x^2*dy-6*x*y*dx = 0
-3*x^2*dy/dx- 6*x*y = 0
as he went thru.
In your work, you didn't use the product rule. TKH came up with the correct final equation except that you then have to put the three pieces together and solve for (dy/dx)
3*x^2 -3*x^2*(dy/dx) - 6*x*y +
2*x*(2*y*(dy/dx)) + 2*y^2 = 0

(-3x^2+4xy)(dy/dx) = -3x^2+6xy-2y^2
etc.