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#