Question

@satellite73

Answer 1

can you please explain each step???

Answer 2

the derivatative of 4 is zero
the derivative of
\[xf(x)\] is
\[f(x)+xf'(x)\] by the product rule, so the derivative of
\[xy\] is
\[y+xy'\]

Answer 3

and by the chain rule, the derivative of
\[f^3(x)\] is
\[3f^2(x)\times f'(x)\] so the derivative of
\[y^3\] is
\[3y^2y'\]

Answer 4

hahaha... it's the first time I see an Asker state that satelite73 is wrong,
@wade123 he is right, not wrong.

Answer 5

Let me explain you what is going on.
As usual, we have y =...some x, right? that is function y with respect to x.

Answer 6

for example, if they say y = 3x +4, you are ok with it, right? but if they say y -3x=4, you feel it is "not good" at all, but actually, they are the same.

Answer 7

no matter what y is, no matter where it is, it is function with respect to x always.

Answer 8

so that, when taking derivative, you treat it as a function,

Answer 9

Now, I don't use \(\dfrac{dy}{dx}\) but y' to indicate \(\dfrac{dy}{dx}\), is it ok to you?

Answer 10

the left hand side first:
\[4' -(xy)'= 0-(x'y +y'x) = -y - xy'\]
got it?

Answer 11

Let me know when you get it.

Answer 12

not finish yet, we just do the left hand side

Answer 13

now, the right hand side
\((y^3)' = 3y^2*y'\) that is the rule. If you don't know, take it from now. Because y is a function respect to x, hence when taking derivative, you must * y' at the end.

Answer 14

Combine the two
\[-y-xy'= 3y^2y'\]

Answer 15

isolate y'

Answer 16

\[3y^2y'+xy'=-x\]

Answer 17

factor y'

Answer 18

\[y'(3y^2+x) = -x\]

Answer 19

then
\[y'=\dfrac{-x}{3y^2+x}\]

Answer 20

Now, replace \(y'=\dfrac{dy}{dx}\)
you have
\[\dfrac{dy}{dx}=\dfrac{-x}{3y^2+x}\]
Got me?

Answer 21

good, :) you are so brave to say " I don't think you are right " to him. I wish I have that manner.