Question

find the implicit derivative of 2xy+y^2=x+y

Answer 1

\[2(y+xy')+2yy'=1+y'\] us a start

Answer 2

why is y' added on to it all?

Answer 3

it is not added for sure
are you asking why it is there?

Answer 4

yes

Answer 5

ok the idea is that in this curve, at least locally, \(y\) is some function of \(x\) i.e. \(y=f(x)\) even though you don't solve for it

Answer 6

so \[ 2xy+y^2=x+y\] is the same as \[ 2xf(x)+f(x)^2=x+f(x)\]

Answer 7

now when you take the derivative, you need the chain rule and also the product rule
for example, the derivative of \[xf(x)\] is
\[f(x)+xf'(x)\] by the product rule

Answer 8

it is easier to write
\[y+xy'\] same thing

Answer 9

the derivative of
\[f^2(x)\] is
\[2f(x)f'(x)\] by the chain rule
here we write the derivative of \(y^2\) is \(2yy'\) again the same thing

Answer 10

if that is not clear imagine that \(y=\sin(x)\) and think of what you would have to do to take the derivative of
\[2x\sin(x)+\sin^2(x)\]

Answer 11

you would need the product and chain rule to do them both
\[\LaTeX\] practice is all

Answer 12

I'm still confused on how you would solve this

Answer 13

d/dx (2xy + y^2 = x + y)
use product rule for 2xy part:
(derivative of the 1st * 2nd as is) + (derivative of the 2nd * 1st as is)
so,
[ d/dx (2x) * y ] + [ d/dx y * (2x) ] = [2*y] + [ (dy/dx) * 2x]
That's for the 2xy part.

derivative of y^2 is 2y dy/dx because you move the exponent to the front of the expression, reduce the exponent by subtracting 1, and stick dy/dx at the end.

derivative of x is just 1 because the exponent is understood to be 1 so put that in the front of the expression, then subtract the exponent of 1 by 1 to get 0, anything to the 0 power is just 1. dx/dx is just 1/1=1 so you don't need to write it

derivative of y is dy/dx...same explanation as d/dx of x but you have to keep dy/dx part.

So, the implicit differentiation for the whole thing is:
2y + 2x(dy/dx) + 2y(dy/dx) = 1 + dy/dx
Then, move all dy/dx to one side and the terms w/o that to the other side:
2x(dy/dx) + 2y (dy/dx) -(dy/dx) = 1 - 2y
Factor out dy/dx:
(dy/dx)[2x + 2y - 1] = 1-2y
Keep just dy/dx part and move the rest over to the other side by division:
dy/dx = (1-2y)/(2x+2y-1)
This is your final answer!