What is the derivative of #arctan(x)/(1+x^2)#?

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Answer 1 · 2017-02-09T14:29:45

#y' = (1 - 2xarctanx)/(1 + x^2)^2#

Let first derive #d/dxarctanx#.

#y= arctanx -> tany = x#

If we differentiate implicitly, we get

#sec^2y(dy/dx) = 1#

#dy/dx = 1/sec^2y#

We now use the pythagorean identity #tan^2theta + 1 = sec^2theta#:

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

Since #x = tany#:

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

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Let's now use the quotient rule to differentiate the entire expression.

#y' = (1/(1 + x^2) * (1 + x^2) - 2xarctanx)/(1 + x^2)^2#

#y' = (1 - 2xarctanx)/(1 + x^2)^2#

Hopefully this helps!