## Calculus: Early Transcendentals 8th Edition

$$\frac{dy}{dx}=\frac{e^{-2x}-2xe^{-2x}}{2\sqrt{1+xe^{-2x}}}$$
$$y=\sqrt{1+xe^{-2x}}$$ $$\frac{dy}{dx}=\frac{d\sqrt{1+xe^{-2x}}}{dx}$$ Let $u=1+xe^{-2x}$ and $y=\sqrt u$. Then, according to the Chain Rule, $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$$ $$\frac{dy}{dx}=\frac{d(\sqrt u)}{du}\frac{d(1+xe^{-2x})}{dx}$$ $$\frac{dy}{dx}=\frac{1}{2\sqrt u}\times(0+\frac{d(x)}{dx}\times e^{-2x}+x\times\frac{d(e^{-2x})}{dx})$$ $$\frac{dy}{dx}=\frac{1}{2\sqrt{1+xe^{-2x}}}\times(e^{-2x}+x\times\frac{d(e^{-2x})}{dx})$$ Here, we also find that we cannot differentiate $e^{-2x}$ right away. So we would carry out the Chain Rule one more time. Let $v=-2x$, then $e^{-2x}=e^v$. According to the Chain Rule, we have $$\frac{d(e^{-2x})}{dx}=\frac{d(e^v)}{dv}\frac{dv}{dx}$$ $$\frac{d(e^{-2x})}{dx}=e^v\times\frac{d(-2x)}{dx}$$ $$\frac{d(e^{-2x})}{dx}=e^v\times(-2)$$ $$\frac{d(e^{-2x})}{dx}=-2e^v=-2e^{-2x}$$ Therefore, $$\frac{dy}{dx}=\frac{1}{2\sqrt{1+xe^{-2x}}}\times(e^{-2x}+x\times(-2e^{-2x}))$$ $$\frac{dy}{dx}=\frac{e^{-2x}-2xe^{-2x}}{2\sqrt{1+xe^{-2x}}}$$