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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?

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    After $y=\tan(\tan^{-1}x)$, then what we get is $$\tan^{-1}y=\tan^{-1}x.$$ The function $\tan^{-1}$ is $1-1$, and so $y=x$, which is a straight line. – 2017-02-25
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    Do you realize that the arctangent function $\arctan(x)=\tan^{-1}(x)$ is the inverse function of the tangent function? Hence $\tan(\arctan(x))=x$. – 2017-02-25

2 Answers 2

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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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$y = tan(tan^{-1}x)$

$y = x$

$dy/dx = 1$

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    hope this helps – 2017-02-25