Find rate of change of implicit function.

Question

There is a function like that : $\frac{y}{x}= a+ bx^2+cy^4$. I want to find rate of change of $\frac{y}{x}$. If I find $\frac{dy}{dx}$ or $\frac{dx}{dy}$ it will give me rate of change of y (or x) respectively to x (or y). But I want to find rate of the change of the $\frac{dy}{dx}$ itself. How can I do it?

Answer 1

$$y=ax+abx^3+cxy^4$$, you need to find $y'$ and $y''$, so d.w.r. t. $x$, $$y'=a+3abx^2+cy^4+4cxy^3 y' \implies (1-4cxy^3)y'=a+3abx^2 \implies y'=\frac{a+3abx^2}{1-4cxy^3}.$$ Again d.w.r.t. $x$, to get $$y''=\frac{(1-4cxy^3)(6abcx)-(a+3abx^2)(-4cy^3-12cxy^2 y')}{(1-4cxy^3)^2}$$

Answer 2

The stages of implicit differentiation may also be handled in this way. Beginning with $$ \frac{d}{dx} \ [ \ \frac{y}{x} \ ] \ \ = \ \ \frac{d}{dx} \ [ \ a+ bx^2+cy^4 \ ] \ \ \Rightarrow \ \ \frac{x·y' \ - \ y}{x^2} \ \ = \ \ 2bx \ + \ 4cy^3 · y' $$ $$ \Rightarrow \ \ x·y' \ - \ y \ \ = \ \ 2bx^3 \ + \ 4cx^2y^3 · y' \quad \quad \mathbf{[ \ A \ ]} \ \ . $$ We can re-arrange this equation to obtain an expression for the first derivative as $$ (x \ - \ 4cx^2y^3) · y' \ \ = \ \ 2bx^3 \ + \ y \ \ \Rightarrow \ \ y' \ \ = \ \ \frac{2bx^3 \ + \ y}{x \ · \ (1 \ - \ 4cxy^3)} \ \ . $$

We will set this aside for the moment to differentiate implicitly equation $ \ \mathbf{A} \ \ , \ $ producing $$ \frac{d}{dx} \ [ \ x·y' \ - \ y \ ] \ \ = \ \ \frac{d}{dx} \ [ \ 2bx^3 \ + \ 4cx^2y^3 · y' \ ] $$ $$ \Rightarrow \ \ ( \ y' \ + \ x·y'' \ ) \ - \ y' \ \ = \ \ 6bx^2 \ + \ 4c·( \ 2xy^3 · y' \ + \ 3·x^2y^2 · y' \ + \ x^2y^3 · y'' \ ) $$ $$ \Rightarrow \ \ x·y'' \ \ = \ \ 6bx^2 \ + \ ( \ 8cxy^3 · y' \ + \ 12cx^2y^2 \ ) · y' \ + \ 4cx^2y^3 · y'' $$ $$ \Rightarrow \ \ x·(1 \ - \ 4cxy^3)·y'' \ \ = \ \ 6bx^2 \ + \ 4cxy^2·( \ 2y \ + \ 3x \ ) · y' \ \ . $$ In this way, we have avoided needing to apply the Quotient Rule to our ratio for $ \ y' \ . $ Instead, we now insert our result for the first derivative into this latest equation to obtain $$ y'' \ \ = \ \ \frac{1}{x·(1 \ - \ 4cxy^3)} \ · \ \ \left[ \ 6bx^2 \ + \ 4cxy^2·( \ 2y \ + \ 3x \ ) · \left(\frac{2bx^3 \ + \ y}{x \ · \ (1 \ - \ 4cxy^3)} \right) \ \right] $$ $$ = \ \ \frac{6bx^2}{x \ · \ (1 \ - \ 4cxy^3)} \ + \ \frac{4cy^2·( \ 2y \ + \ 3x \ ) · ( 2bx^3 \ + \ y)}{x·(1 \ - \ 4cxy^3)^2} $$ $$ = \ \ \frac{6bx^2 · (1 \ - \ 4cxy^3) \ + \ 4cy^2·( \ 2y \ + \ 3x \ ) · ( 2bx^3 \ + \ y)}{x·(1 \ - \ 4cxy^3)^2} $$

$$ = \ \ \frac{24bcx^4y^2 \ - \ 8bcx^3y^3 \ + \ 6bx^2 \ + \ 12cxy^3 \ + \ 8cy^4}{x·(1 \ - \ 4cxy^3)^2} \ \ . $$