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Implicit differentiation question/problem

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How would you find

dy

dx

for 2xy2 = x2y+3

I know that one of the first steps of the working is

2y2 + 4xydy = 2xy + x2dy

...............dx..................dx

I know how to go from there but I dont know how you would get there.. and why?

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You get there by using the product rule. I can't answer why :D , I'm not here to derive the way product rule works :lol:

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dy/ dx for the above question is:

2xy - 2y^2

--------------

4xy - x^2

The produce rule is: (1st x Differential of the 2nd )+ (2nd x differential of the 1st)

Edit: I just saw your working.. Here is how to do the rest.

Bring all those with dy/dx to the left hand side..

so...

2y^2 + 4xy(dy/dx) = 2xy + x^2(dy/dc)

goes to:

4xy (dy/dx) - x^2(dy/dx) = 2xy - 2y^2.

Then factor out the dy/dx:

DY/DX{4xy - x^2} = 2xy - 2y^2

Then divide by 4xy - x^2

DY/DX = 2xy - 2y^2

--------------

4xy - x^2

Edited by uggy

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Do something to make your life easier, do the problem as if it was a regular differentiation, but every time you take the derivative of y just write dy/dx next to it:

xy = 2

you dont notice, but when you take the derivative of x the answer isnt 1, its 1*dx/dx which is one. So for xy = 2 just do a product rule:

(dx/dx)y + x (dy/dx) = 0

see all you do is put in dy/dx whenever you TAKE THE DERIVATE OF Y.

As far as how to do the rest uggy took care of most of it. See in reality what you have is:

2xy2 = x2y+3

(2)(dx/dx)y^2 + 2x(2y)(dy/dx) = (2x)(dx/dx):blink: + 2x(dy/dx) + 0, which is the same as 2y^2 + 4xy(dy/dx) = 2xy+ 2x(dy/dx)

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