Help me understand the differentiation of $f_{x}+f_{y}\frac{dy}{dx}=0$

Question

Help me understand the differentiation of $f_{x}+f_{y}\frac{dy}{dx}=0$

We are learning singular points in college. There is a subtopic, rather a derivation in the course material where:

Differentiating $f_{x}+f_{y}\frac{dy}{dx}$ w.r.t $x$ we have,

$f_{xx}+f_{xy}\frac{dy}{dx}+ [(f_{yx}+f_{yy}\frac{dy}{dx})\frac{dy}{dx}]+f_{y}\frac{d^{2}y}{dx^{2}}=0$

I am not able to understand how did they get the derivation. To be precise I am not able to understand how did they get the term inside the square brackets.

the terms with subscript in the equation is partial derivation

Thanks in advance!

Answer 1

Hint. By differentiating the product $f_{y}\frac{dy}{dx}$ we obtain, $$\frac{d}{dx}\left(f_{y}\frac{dy}{dx}\right)= \frac{d}{dx}\left(f_{y}\right)\frac{dy}{dx}+f_{y}\frac{d^{2}y}{dx^{2}}.$$ Now what is $\frac{d}{dx}\left(f_{y}\right)$? Note that here you need the total-derivative, $$\frac{d}{dx}\left(\cdot\right)=\frac{\partial}{\partial x}(\cdot)+\frac{\partial}{\partial y}(\cdot)\cdot \frac{dy}{dx}$$ If you use it with $f_y$, you get $$\frac{d}{dx}\left(f_y\right)=\frac{\partial}{\partial x}(f_y)+\frac{\partial}{\partial y}(f_y)\cdot \frac{dy}{dx}=f_{yx}+f_{yy}\frac{dy}{dx}$$ which is exactly the term inside the square brackets. In a similar way you obtain the first terms, $$\frac{d}{dx}\left(f_x\right)=\frac{\partial}{\partial x}(f_x)+\frac{\partial}{\partial y}(f_x)\cdot \frac{dy}{dx}=f_{xx}+f_{xy}\frac{dy}{dx}.$$

Answer 2

NOTE:$\dfrac{d}{dx}f(x,y)=f_x+f_y\dfrac{dy}{dx}$

Now ,$\dfrac{d}{dx}\{f_x+f_y\dfrac{dy}{dx}\}=\dfrac{d}{dx}(f_x)+\dfrac{d}{dx}(f_y\dfrac{dy}{dx})$

$=(f_{xx}+f_{xy}\dfrac{dy}{dx})+\{(\dfrac{d}{dx}f_y)\times \dfrac{dy}{dx})+f_y\dfrac{d}{dx}(\dfrac{dy}{dx})\}$

$=(f_{xx}+f_{xy}\dfrac{dy}{dx})+\{f_{yx}+f_{yy}\dfrac{dy}{dx}\}\dfrac{dy}{dx}+f_y\dfrac{d^2y}{dx^2}$