## Calculus: Early Transcendentals 8th Edition

$$\frac{dy}{dx}=\frac{-e^x}{2\sqrt{2-e^x}}$$
$$y=\sqrt{2-e^x}$$ $$y=(2-e^x)^{1/2}$$ $$\frac{dy}{dx}=\frac{d(2-e^x)^{1/2}}{dx}$$ Take $u=2-e^x$ and $y=u^{1/2}$. Then, according to Chain Rule, $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$$ $$\frac{dy}{dx}=\frac{d(u^{1/2})}{du}\frac{d(2-e^x)}{dx}$$ $$\frac{dy}{dx}=\frac{1}{2}u^{-1/2}\times(-e^x)$$ $$\frac{dy}{dx}=\frac{1}{2\sqrt u}\times (-e^x)$$ $$\frac{dy}{dx}=\frac{-e^x}{2\sqrt{2-e^x}}$$
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