Question
Differentiate the following w.r.t. x:
$\tan^{-1}(\sec\text{x}+\tan\text{x}),\frac{-\pi}{2}<\text{x}<\frac{\pi}{2}$

Answer

Let $\text{y}=\tan^{-1}(\sec\text{x}+\tan\text{x})$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\tan^{-1}(\sec\text{x}+\tan\text{x})$
$=\frac{1}{1+(\sec\text{x}+\tan\text{x})^2}\cdot\frac{\text{d}}{\text{dx}}(\sec\text{x}+\tan\text{x})$
$=\frac{1}{1+\sec^2\text{x}+\tan^2\text{x}+2\sec\text{x}\cdot\tan\text{x}}\big[\sec\text{x}\cdot\tan\text{x}+\sec^2\text{x}\big]$
$=\frac{1}{\big(\sec^2\text{x}+\sec^2\text{x}+2\sec\text{x}\cdot\tan\text{x}\big)}\cdot\sec\text{x}\cdot(\sec\text{x}+\tan\text{x})$
$=\frac{1}{2\sec\text{x}(\tan\text{x}+\sec\text{x})}\cdot\sec\text{x}(\sec\text{x}+\tan\text{x})=\frac{1}{2}$

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