<s>[INST] How to prove that graph of $\tan (\tan^{-1} x) $is a straight line? 
How to prove that graph of $\tan (\tan^{-1} x) $is a straight line ?

$y=\tan(\tan^{-1} x) \Rightarrow y'=\sec^2(\tan^{-1} x)\cdot \frac{1}{x^2+1}$
but how can $y'=1$ OR = some constant $k$ be proved?
what I did is:
$y'=\left[\tan^2(\tan^{-1}x)+1\cdot \frac{1} {x^2+1}\right]=\left[(\tan(\tan^{-1}x))^2+1)
\cdot \frac{1} {x^2+1}\right]=\frac{y^2+1}{ x^2+1}$
What next?
 [/INST] Note that $\sec(\arctan(x))=\sqrt{x^2+1}$.  Hence, 
$$\begin{align}
\frac{d}{dx}\tan(\arctan(x))&=\sec^2(\arctan(x))\left(\frac{d}{dx}\arctan(x)\right)\\\\
&=\left(\sqrt{x^2+1}\,\,\right)^2\,\left(\frac{1}{1+x^2}\right)\\\\
&=1
\end{align}$$
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