What is the derivative of $\frac{arctan (x)}{1 + x^{2}}$ $arctan(x)/(1+x^2)$ ?

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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 $x = tany$ :

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

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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!

What is the derivative of arctan(x)/(1+x^2)arctan(x)1+x2arctan(x)/(1+x^2)?