How do you find the derivative of $tan^-1(x^2)$ ?

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Answer 1 · 2015-07-05T03:11:15

Assuming you don't remember the derivative of $arctanu$ :

$y = arctanx^2$

$tany = x^2$

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

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

$= (2x)/(1+tan^2y)$

Since $tany = x^2$ :

$= color(blue)((2x)/(1+x^4)$

If you did remember it:

$d/(dx)[arctanu] = 1/(1+u^2)((du)/(dx))$

$u = x^2$

$=> 1/(1+(x^2)^2)*2x = color(darkblue)((2x)/(1+x^4))$

How do you find the derivative of tan^-1(x^2)tan^-1(x^2)?