How do you find the derivative of #arctan (x/2)#?

Question

1283 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-18T17:26:12

By definition, we know that if #y=arctan(x)#, then

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

Here, applying the chain rule, we can solve this problem naming #u=x/2#. Just remembering the chain rule definition:

#dy/(dx)=dy/(du)*(du)/(dx)#

#dy/(du)=(u')/(1+u^2)#

#(du)/(dx)=1/2#

#dy/(dx)=(u')/(2(1+u^2))#

Substituing #u#, we get

#dy/(dx)=(1/2)/(1+((x^2)/4))#=#(1/2)/((4+x^2)/4)#=#4/(2(4+x^2))#

#dy/(dx)=2/(4+x^2)#