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)#