Using implicit differentiation with a fraction

Question

How do I solve this? What steps? I have been beating my head into the wall all evening.

$$ x^2 + y^2 = \frac{x}{y} + 4 $$

Answer 1

We have:

$$ \frac{d(x^2)}{dx} + \frac{d(y^2)}{dy}\cdot \frac{dy}{dx} = \frac{d( \frac{x}{y})}{dx} + \frac{d(4)}{dx}$$

$$2x+2y\frac{dy}{dx} = \frac{y-x \frac{dy}{dx} }{ y^2} +0$$

Can you finish?

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Okay, to finish it up:

$$2y\frac{dy}{dx} - \frac{y-x \frac{dy}{dx} }{ y^2} = -2x $$
Multiplying both sides by $y^2$ $$2y^3\frac{dy}{dx}-(y-x\frac{dy}{dx})=-2xy^2$$ $$2y^3\frac{dy}{dx}+x\frac{dy}{dx}=-2xy^2 +y$$ $$\frac{dy}{dx} \cdot (2y^3+x) =-2xy^2 +y$$ $$\frac{dy}{dx} = \frac{-2xy^2 +y}{2y^3+x}$$

Answer 2

We can multiply both sides of the equation $$x^2 + y^2 = \frac{x}{y} + 4 \tag{1}$$ by $y$ to obtain $$x^2y + y^3 = x + 4y \tag{2}$$ Differentiating equation 2 implicitly with respect to $x$ yields \begin{align*} 2xy + x^2y' + 3y^2y' & = 1 + 4y'\\ (x^2 + 3y^2 - 4)y' & = 1 - 2xy\\ y' & = \frac{1 - 2xy}{x^2 + 3y^2 - 4} \tag{3} \end{align*} Let's see why this answer is equivalent to that given by @CivilSigma. If we solve equation 1 for $x/y$, we obtain $$\frac{x}{y} = x^2 + y^2 - 4$$ Hence, \begin{align*} y' & = \frac{1 - 2xy}{x^2 + 3y^2 - 4}\\ & = \frac{1 - 2xy}{x^2 + y^2 - 4 + 2y^2}\\ & = \frac{1 - 2xy}{\dfrac{x}{y} + 2y^2}\\ & = \frac{y - 2xy^2}{x + 2xy^3} \end{align*}