What's the derivative of $arctan(x/a)$ ?

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Answer 1 · 2016-08-06T08:03:02

$d/(dx)arctan(x/a)=a/(x^2+a^2)$

$y=arctan(x/a)$ is equivalent to $tany=x/a$

Now taking derivative of both sides

$sec^2yxx(dy)/(dx)=1/a$ or

$(dy)/(dx)=1/axx1/sec^2y$

= $1/axx1/(1+tan^2y)$

= $1/axx1/(1+x^2/a^2)$

= $1/axxa^2/(x^2+a^2)$

= $a/(x^2+a^2)$

What's the derivative of arctan(x/a)arctan(x/a)?