Question

evaluate the derivative of the following function.

$f(x)=2\csc^{-1}(\tan e^{x})$

$\frac{d}{dx}f(x)=\square$

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:

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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 \(\tan t=\frac{\sin t}{\cos t}\)).

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Answer:

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