## University Calculus: Early Transcendentals (3rd Edition)

$$\frac{dy}{dx}=\frac{2-2xye^{x^2y}}{x^2e^{x^2y}-2}$$
$$e^{x^2y}=2x+2y$$ 1) Differentiate both sides of the equation with respect to $x$: $$\frac{d}{dx}(e^{x^2y})=\frac{d}{dx}(2x+2y)$$ $$e^{x^2y}\frac{d}{dx}(x^2y)=2+2\frac{dy}{dx}$$ $$e^{x^2y}(2xy+x^2\frac{dy}{dx})=2+2\frac{dy}{dx}$$ $$2xye^{x^2y}+x^2e^{x^2y}\frac{dy}{dx}=2+2\frac{dy}{dx}$$ 2) Collect all the terms with $dy/dx$ onto one side and solve for $dy/dx$: $$x^2e^{x^2y}\frac{dy}{dx}-2\frac{dy}{dx}=2-2xye^{x^2y}$$ $$\frac{dy}{dx}=\frac{2-2xye^{x^2y}}{x^2e^{x^2y}-2}$$