How Do You Derive the Slope from the Equation 2xy^2+xy=y?

[Tom McCurdy]

taking derivative of 2xy^2+xy=y

split up to using product rule

2x----2
y^2---2y\frac{dy}{dx}

4yx\frac{dy}{dx}+2y^2+y+x\frac{dy}{dx}=\frac{dy}{dx}

\frac{dy}{dx} (4x+x-1)=-2y^2-y

\frac{dy}{dx}= \frac{-2y^2-y}{4xy+x-1}

i am trying to figure out the slope of the equation when y-1

am i doing all this for nothing?

 

[Falnom]

Do you mean when y=1?
If so then all you have to do is plug back in y in the orignal equation and solve for x. And then plug in both x and y into your final differential equation. I think that the final slope will then be -9/2.

 

[wais]

No, you are not doing this for nothing. By taking the derivative of the given equation, you are finding the slope of the equation at a specific point (y=1). This information can be useful in solving problems related to the motion or behavior of objects in physics. Additionally, understanding the concept of taking derivatives is important in understanding the underlying principles of physics, such as acceleration and velocity. So, keep up the good work!