What is the second derivative of inverse tangent?

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Answer 1 · 2018-03-03T17:33:40

$\frac{d^{2}}{{d x}^{2}} arctan x = - \frac{2 x}{{(1 + x^{2})}^{2}}$ $d^2/(dx^2) arctanx= -(2x)/(1+x^2)^2$

Let:

$y = arctan x$ $y = arctanx$

so that:

$x = tan y$ $x=tany$

differentiate this last equality with respect to $x$ $x$ :

$1 = {sec}^{2} y \frac{d y}{d x}$ $1= sec^2y dy/dx$

Now using the trigonometric inequality:

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

we have:

$1 = (1 + {tan}^{2} y) \frac{d y}{d x}$ $1 = (1+tan^2y)dy/dx$

$1 = (1 + x^{2}) \frac{d y}{d x}$ $1 =(1+x^2)dy/dx$

that is:

$\frac{d y}{d x} = \frac{1}{1 + x^{2}}$ $dy/dx =1/(1+x^2)$

Differentiate again using the chain rule:

$\frac{d^{2} y}{{d x}^{2}} = \frac{d}{d x} \frac{1}{1 + x^{2}}$ $(d^2y)/(dx^2) = d/dx1/(1+x^2)$

$\frac{d^{2} y}{{d x}^{2}} = - \frac{1}{{(1 + x^{2})}^{2}} \frac{d}{d x} (1 + x^{2})$ $(d^2y)/(dx^2) = -1/(1+x^2)^2 d/dx (1+x^2)$

$\frac{d^{2} y}{{d x}^{2}} = - \frac{2 x}{{(1 + x^{2})}^{2}}$ $(d^2y)/(dx^2) = -(2x)/(1+x^2)^2$