Question

implicit derivative of y^2-xy-3x=1

Answer 1

Well, do it.

Answer 2

i think that it's 2yy'-x'y* -xy' -3x=0

Answer 3

is that right?

Answer 4

Not bad. One piece at a time.

\(\dfrac{d}{dx}y^{2} = 2yy'\)

\(\dfrac{d}{dx}-xy = -(xy' + y)\) ==> x' = 1 :-)

\(\dfrac{d}{dx}-3x = -3\) ==> You didn't do anything to this piece.

\(\dfrac{d}{dx}1 = 0\)

Answer 5

i forgot product rule is f'g+fg'

Answer 6

i thought it was f'g * fg'

Answer 7

why do we take derivative of 3x though? I thought we were just taking the derivative of y

Answer 8

so then y'=-1

Answer 9

thanks a ton

Answer 10

Have to get them all. Good work.

2yy' - xy' - y - 3 = 0

y'(2y - x) = y+3

\(y' = \dfrac{y+3}{2y-x}\)

How did you get -1 out of that?