What is the derivative of $f(x)=tan^-1(x)$ ?

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Answer 1 · 2015-07-01T16:47:50

I seem to recall my professor forgetting how to deriving this. This is what I showed him:

$y = arctanx$

$tany = x$

$sec^2y (dy)/(dx) = 1$

$(dy)/(dx) = 1/(sec^2y)$

Since $tany = x/1$ and $sqrt(1^2 + x^2) = sqrt(1+x^2)$ , $sec^2y = (sqrt(1+x^2)/1)^2 = 1+x^2$

$=> color(blue)((dy)/(dx) = 1/(1+x^2))$

I think he originally intended to do this:

$(dy)/(dx) = 1/(sec^2y)$

$sec^2y = 1+tan^2y$

$tan^2y = x -> sec^2y = 1+x^2$

$=> (dy)/(dx) = 1/(1+x^2)$

What is the derivative of f(x)=tan^-1(x)f(x)=tan^-1(x) ?