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

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Answer 1 · 2017-05-13T14:26:56

The derivative is $=-(2x)/(x^4+2x^2+2)$

We need

$(tanx)'=sec^2x$

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

$(1/x)'=-1/x^2$

Let $y=arctan(1/(1+x^2))$

Then,

$tany=1/(1+x^2)$

Differentiating both sides

$sec^2y*dy/dx=(-2x)/(1+x^2)^2$

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

Therefore,

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

$=-(2x)/(x^4+2x^2+2)$

How do you find the derivative of arctan^-1 (1/(1+x^2))arctan^-1 (1/(1+x^2))?