How to find dy/dx = - fx/fy?

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Solution 1

Take the differential of the given equation $3x^2-y^2+x^3=0$:

$$6x\,dx-2y\,dy+3x^2\,dx=0.$$

This gives you the requested derivative,

$$\frac{dy}{dx}=\frac{6x+3x^2}{2y}.$$

Solution 2

This is coming from total and implicit differentiation. Suppose that you have a function $$F(x,y)=0$$ then the total derivative write $$dF(x,y)=\frac{\partial F(x,y)}{\partial x}dx+\frac{\partial F(x,y)}{\partial y}dy=0$$ and what you want is $\frac{dy}{dx}$. So, $$\frac{dy}{dx}=-\frac{\frac{\partial F(x,y)}{\partial x}}{\frac{\partial F(x,y)}{\partial y}}$$

In the example you give, you lost parentheses for the last expression.

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Updated on May 10, 2020

Comments

  • HQuser
    HQuser over 3 years

    I need some walkthrough in solving the following question:

    find dy/dx = - fx/fy?
    3x^2 - y^2 + x^3 = 0.
    

    I need to know the method to solve this question.

    According to my understanding what I have concluded that:

    1. Take partial derivative of the question w.r.t. x.

    2. Take partial derivative of the question w.r.t. y.

    3. Put the values of both in the equation: -fx/fy and simplify.

    Like:

    fx = 6x + 3x^2
    fy = -2y
    -fx/fy = -(6x+3x^2)/(-2y) 
    = (6x + 3x^2)/2y
    

    I wanted to know if my conclusion is correct or there's something else which I am missing? Thanks!

  • HQuser
    HQuser over 8 years
    I didn't still got it, Do I have to find Fx and Fy and then again the derivative of -Fx/Fy?
  • HQuser
    HQuser over 8 years
    So this means I have to prove that L.H.S = R.H.S.?
  • Admin
    Admin over 8 years
    That's the solution.