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

The tangent line to the curve at $P(1,0)$ is $$y=\frac{1}{2}x-\frac{1}{2}$$
$$x\sqrt{1+2y}+y=x^2$$ $$x(1+2y)^{1/2}+y=x^2$$ The graph of the curve is enclosed below. a) As the point $(1,0)$ lies in the curve, $P$ satisfies the equation. b) Find $dy/dx$ using implicit differentiation: $$(1+2y)^{1/2}+x\frac{d}{dx}(1+2y)^{1/2}+\frac{dy}{dx}=2x$$ $$(1+2y)^{1/2}+x\Big(\frac{1}{2}(1+2y)^{-1/2}\frac{d}{dx}(1+2y)\Big)+\frac{dy}{dx}=2x$$ $$(1+2y)^{1/2}+\frac{x}{2(1+2y)^{1/2}}(2\frac{dy}{dx})+\frac{dy}{dx}=2x$$ $$\frac{x}{(1+2y)^{1/2}}\frac{dy}{dx}+\frac{dy}{dx}=2x-(1+2y)^{1/2}$$ $$\frac{dy}{dx}\Big(\frac{x}{(1+2y)^{1/2}}+1\Big)=2x-(1+2y)^{1/2}$$ $$\frac{dy}{dx}\Big(\frac{x+(1+2y)^{1/2}}{(1+2y)^{1/2}}\Big)=2x-(1+2y)^{1/2}$$ $$\frac{dy}{dx}=\frac{2x-(1+2y)^{1/2}}{\frac{x+(1+2y)^{1/2}}{(1+2y)^{1/2}}}=\frac{(2x-(1+2y)^{1/2})(1+2y)^{1/2}}{x+(1+2y)^{1/2}}$$ $$\frac{dy}{dx}=\frac{(2x-\sqrt{1+2y})\sqrt{1+2y}}{x+\sqrt{1+2y}}$$ - For $P(1,0)$: $$\frac{dy}{dx}=\frac{(2\times1-\sqrt{1+2\times0})(\sqrt{1+2\times0})}{1+\sqrt{1+2\times0}}=\frac{(2-\sqrt1)\sqrt1}{1+\sqrt1}$$ $$\frac{dy}{dx}=\frac{1\times1}{2}=\frac{1}{2}$$ c) The tangent line to the curve at $P(1,0)$ is $$y-0=\frac{1}{2}(x-1)$$ $$y=\frac{1}{2}x-\frac{1}{2}$$