Derivative of $$$\tan{\left(x \right)} \sec{\left(x \right)}$$$
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Solution
Apply the product rule $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ with $$$f{\left(x \right)} = \sec{\left(x \right)}$$$ and $$$g{\left(x \right)} = \tan{\left(x \right)}$$$:
$${\color{red}\left(\frac{d}{dx} \left(\tan{\left(x \right)} \sec{\left(x \right)}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(\sec{\left(x \right)}\right) \tan{\left(x \right)} + \sec{\left(x \right)} \frac{d}{dx} \left(\tan{\left(x \right)}\right)\right)}$$The derivative of the tangent is $$$\frac{d}{dx} \left(\tan{\left(x \right)}\right) = \sec^{2}{\left(x \right)}$$$:
$$\tan{\left(x \right)} \frac{d}{dx} \left(\sec{\left(x \right)}\right) + \sec{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\tan{\left(x \right)}\right)\right)} = \tan{\left(x \right)} \frac{d}{dx} \left(\sec{\left(x \right)}\right) + \sec{\left(x \right)} {\color{red}\left(\sec^{2}{\left(x \right)}\right)}$$The derivative of the secant is $$$\frac{d}{dx} \left(\sec{\left(x \right)}\right) = \tan{\left(x \right)} \sec{\left(x \right)}$$$:
$$\tan{\left(x \right)} {\color{red}\left(\frac{d}{dx} \left(\sec{\left(x \right)}\right)\right)} + \sec^{3}{\left(x \right)} = \tan{\left(x \right)} {\color{red}\left(\tan{\left(x \right)} \sec{\left(x \right)}\right)} + \sec^{3}{\left(x \right)}$$Simplify:
$$\tan^{2}{\left(x \right)} \sec{\left(x \right)} + \sec^{3}{\left(x \right)} = \left(-1 + \frac{2}{\cos^{2}{\left(x \right)}}\right) \sec{\left(x \right)}$$Thus, $$$\frac{d}{dx} \left(\tan{\left(x \right)} \sec{\left(x \right)}\right) = \left(-1 + \frac{2}{\cos^{2}{\left(x \right)}}\right) \sec{\left(x \right)}$$$.
Answer
$$$\frac{d}{dx} \left(\tan{\left(x \right)} \sec{\left(x \right)}\right) = \left(-1 + \frac{2}{\cos^{2}{\left(x \right)}}\right) \sec{\left(x \right)}$$$A