Question

whats the derivative of 2x - y^2?

Answer 1

2-2y

Answer 2

??? how?

Answer 3

you have no equation here

Answer 4

dy/dx=2x-y^2

Answer 5

sorry your right.

Answer 6

if it is
\[2x-y^2=0\] for example then you can find \(y'\) or \(x'\)

Answer 7

i know its y''=2-... i have no idea how to differentiate y^2 with respect to x

Answer 8

@Rezz5 we need a better starting point

Answer 9

deravative of 2x = 2x^(1-1) = 2x^0 and x^0 =1 so 2*1=2
deravative of y^2 = 2*y^(2-1) = 2y^1 =2y

Answer 10

@Sjano it cant be 2y,

Answer 11

\[2x-y^2\] is just an expression
it does not define y as a function of x, or x as a function of y
it is not an equation at all

Answer 12

okay?

Answer 13

it is not a curve, it is not anything.

Answer 14

but whats is its derivative?

Answer 15

would it be y''=2-2y'y?

Answer 16

it doesn't have one because y is not a function of x, and x is not a function of y
you need an EQUATION of some kind
is this a problem from a book?

Answer 17

you are given y'=2x-y^2
i need y''

Answer 18

ok now you have an equation right? with an equal sign

Answer 19

i gave typed it in when you told there was no equation above...

Answer 20

ok now i see it. i didn't see the \(\frac{dy}{dx}=2x-y^2\) so we are thinking that y is some function of x so it is like having
\[f'(x)=2x-f(x)^2\] and we want
\[f''(x)\] take the derivative with respect to x and get
\[f''(x)=2-2f(x)f'(x)\] or
\[y''=2-2yy'\]

Answer 21

now we know
\[y'=2x-y^2\] so we can replace the \(y'\) by \(2x-y^2\) above and get
\[y''=2-2y(2x-y^2)\]

Answer 22

so would the derivative of y^3 with respect to x be (3y^2)*y'?

Answer 23

yea, exactly

Answer 24

THANK YOU.. i remember know!!

Answer 25

yw

Answer 26

now*