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Soroban · #4 May 11th, 2008, 11:07 PM

Hello, chousta!

Quote:

2) The equation: . $-4xy - \frac{2}{xy} - \frac{2}{x^2y^2} \:=\:-6$ .implicitly defines $y$ as a function of $x.$
Find $\frac{dy}{dx}$

Multiply by -1: . $4xy + 2x^{-1}y^{-1} + 2x^{-2}y^{-2} \;=\;6$

Differentiate implicitly: . $4x\frac{dy}{dx} + 4y - 2x^{-1}y^{-2}\frac{dy}{dx} - 2x^{-2}y^{-1} - 4x^{-2}y^{-3}\frac{dy}{dx} - 4x^{-3}y^{-2} \;=\;0$

Rearrange terms: . $4x\frac{dy}{dx} - \frac{2}{xy^2}\frac{dy}{dx} - \frac{4}{x^2y^3}\frac{dy}{dx}\;=\;-4y + \frac{2}{x^2y} + \frac{4}{x^3y^2}$

Multiply by $\frac{x^3y^3}{2}\!:\quad 2x^4y^3\frac{dy}{dx} - x^2y\frac{dy}{dx} - 2x\frac{dy}{dx} \:=\:-2x^3y^4 + xy^2 + 2y$

Factor: . $x(2x^3y^3 - xy - 2)\frac{dy}{dx} \;=\;\text{-}y(2x^3y^3 - xy - 2)$

Therefore: . $\frac{dy}{dx}\;=\; \frac{\text{-}y(2x^3y^3 - xy - 2)}{x(2x^3y^3 - xy - 2)} \;=\;\boxed{-\frac{y}{x}}$