What is the second derivative of inverse tangent?

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Answer 1 · 2018-03-03T17:33:40

$d^2/(dx^2) arctanx= -(2x)/(1+x^2)^2$

Let:

$y = arctanx$

so that:

$x=tany$

differentiate this last equality with respect to $x$ :

$1= sec^2y dy/dx$

Now using the trigonometric inequality:

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

we have:

$1 = (1+tan^2y)dy/dx$

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

that is:

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

Differentiate again using the chain rule:

$(d^2y)/(dx^2) = d/dx1/(1+x^2)$

$(d^2y)/(dx^2) = -1/(1+x^2)^2 d/dx (1+x^2)$

$(d^2y)/(dx^2) = -(2x)/(1+x^2)^2$