evaluate the derivative of the following function.\n\n$f(x)=2\\csc^{-1}(\\tan e^{x})$\n\n$\\frac{d}{dx}f(x)=\…

Explanation:

Step1: Apply the chain rule

The chain rule states that if (y = f(g(x))), then (y^\prime=f^\prime(g(x))\cdot g^\prime(x)). Let (u = \tan(e^{x})), so (y = 2\csc^{- 1}(u)). First, find the derivative of (y) with respect to (u): (\frac{dy}{du}=-\frac{2}{|u|\sqrt{u^{2}-1}}) (since the derivative of (\csc^{-1}(u)) is (-\frac{1}{|u|\sqrt{u^{2}-1}})).

Step2: Find the derivative of (u) with respect to (x)

Now, find the derivative of (u=\tan(e^{x})) with respect to (x). Using the chain rule again, let (v = e^{x}), so (u=\tan(v)). The derivative of (\tan(v)) with respect to (v) is (\sec^{2}(v)), and the derivative of (v = e^{x}) with respect to (x) is (e^{x}). So (\frac{du}{dx}=\sec^{2}(e^{x})\cdot e^{x}).

Step3: Combine the derivatives using the chain rule

By the chain rule (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}). Substitute (\frac{dy}{du}=-\frac{2}{|u|\sqrt{u^{2}-1}}) and (\frac{du}{dx}=\sec^{2}(e^{x})\cdot e^{x}) with (u = \tan(e^{x})) into the formula: [ \begin{align*} \frac{dy}{dx}&=-\frac{2}{|\tan(e^{x})|\sqrt{\tan^{2}(e^{x})-1}}\cdot\sec^{2}(e^{x})\cdot e^{x}\ \end{align*} ] Since (\tan(e^{x})>0) for appropriate domains (assuming (e^{x}) is in the domain where (\tan) is positive and (\csc^{-1}(\tan(e^{x}))) is defined), (|\tan(e^{x})|=\tan(e^{x})). Also, (\sec^{2}(t)=1 + \tan^{2}(t)), so (\sec^{2}(e^{x})=\frac{1}{\cos^{2}(e^{x})}) and (\sqrt{\tan^{2}(e^{x})-1}=\frac{\sqrt{\sin^{2}(e^{x})-\cos^{2}(e^{x})}}{\cos(e^{x})}) (using (\tan t=\frac{\sin t}{\cos t})). [ \begin{align*} \frac{dy}{dx}&=-\frac{2e^{x}\sec^{2}(e^{x})}{\tan(e^{x})\sqrt{\tan^{2}(e^{x})-1}}\ &=-\frac{2e^{x}}{\sin(e^{x})\sqrt{\sin^{2}(e^{x})-\cos^{2}(e^{x})}} \end{align*} ]

Answer:

(-\frac{2e^{x}}{\sin(e^{x})\sqrt{\sin^{2}(e^{x})-\cos^{2}(e^{x})}})