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

$$\frac{d^2y}{dx^2}=\frac{(1+2x^2)e^{x^2}y^2-(xe^{x^2}+1)^2}{y^3}$$
$$y^2=e^{x^2}+2x$$ 1) Find $dy/dx$: Differentiate both sides of the equation with respect to $x$: $$\frac{d}{dx}(y^2)=\frac{d}{dx}(e^{x^2})+\frac{d}{dx}(2x)$$ $$2y\frac{dy}{dx}=e^{x^2}\frac{d}{dx}(x^2)+2$$ $$2y\frac{dy}{dx}=2xe^{x^2}+2$$ $$y\frac{dy}{dx}=xe^{x^2}+1$$ Solve for $dy/dx$: $$\frac{dy}{dx}=\frac{xe^{x^2}+1}{y}$$ 2) Find $d^2y/dx^2$: Differentiate the derivative $dy/dx$ as normal, using Quotient Rule: $$\frac{d^2y}{dx^2}=\frac{d}{dx}\Big(\frac{xe^{x^2}+1}{y}\Big)=\frac{\Big(\frac{d}{dx}(xe^{x^2})+0\Big)y-(xe^{x^2}+1)\frac{dy}{dx}}{y^2}$$ $$\frac{d^2y}{dx^2}=\frac{(e^{x^2}+x\frac{d}{dx}(e^{x^2}))y-(xe^{x^2}+1)\frac{dy}{dx}}{y^2}$$ $$\frac{d^2y}{dx^2}=\frac{(e^{x^2}+x\times(2xe^{x^2}))y-(xe^{x^2}+1)\frac{dy}{dx}}{y^2}$$ $$\frac{d^2y}{dx^2}=\frac{(e^{x^2}+2x^2e^{x^2})y-(xe^{x^2}+1)\frac{dy}{dx}}{y^2}$$ Now we can use the result $dy/dx=\frac{xe^{x^2}+1}{y}$ from part 1) to do substitution here: $$\frac{d^2y}{dx^2}=\frac{(e^{x^2}+2x^2e^{x^2})y-(xe^{x^2}+1)\frac{xe^{x^2}+1}{y}}{y^2}$$ $$\frac{d^2y}{dx^2}=\frac{\frac{(e^{x^2}+2x^2e^{x^2})y^2-(xe^{x^2}+1)^2}{y}}{y^2}$$ $$\frac{d^2y}{dx^2}=\frac{(e^{x^2}+2x^2e^{x^2})y^2-(xe^{x^2}+1)^2}{y^3}$$ $$\frac{d^2y}{dx^2}=\frac{(1+2x^2)e^{x^2}y^2-(xe^{x^2}+1)^2}{y^3}$$