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)/2yI wanted to know if my conclusion is correct or there's something else which I am missing? Thanks!
$\endgroup$2 Answers
$\begingroup$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.
$\endgroup$ 1 $\begingroup$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,
$$\frac{dy}{dx}=\frac{6x+3x^2}{2y}.$$
$\endgroup$ 2
