Easy
Differentiate the following with respect to $x:$ $\cos^{-1}(e^x)$
(N/A) Let $y = \cos^{-1}(e^x).$
Using the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\cos^{-1}(e^x))$
Since $\frac{d}{dx}(\cos^{-1}(u)) = \frac{-1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$,where $u = e^x$:
$\frac{dy}{dx} = \frac{-1}{\sqrt{1-(e^x)^2}} \cdot \frac{d}{dx}(e^x)$
$\frac{dy}{dx} = \frac{-1}{\sqrt{1-e^{2x}}} \cdot e^x$
$\frac{dy}{dx} = \frac{-e^x}{\sqrt{1-e^{2x}}}$
Using the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\cos^{-1}(e^x))$
Since $\frac{d}{dx}(\cos^{-1}(u)) = \frac{-1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$,where $u = e^x$:
$\frac{dy}{dx} = \frac{-1}{\sqrt{1-(e^x)^2}} \cdot \frac{d}{dx}(e^x)$
$\frac{dy}{dx} = \frac{-1}{\sqrt{1-e^{2x}}} \cdot e^x$
$\frac{dy}{dx} = \frac{-e^x}{\sqrt{1-e^{2x}}}$
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