How do you differentiate #Arctan ((x-1)/(x+1))#?

Question

5190 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-14T13:16:56

# 1/(1+x^2), x!=-1, x in RR.#

Let, #y=arc tan{(x-1)/(x+1)}#

Clearly, #x in RR-{-1}.#

Recall that the Range of #tan# fun. is #RR,# so, we can, suppose,

#x=tantheta," so that, "theta =arc tanx.#

#"Also, "x in RR-{-1} rArr theta in RR-{npi-pi/4 : n in ZZ}.#

Now, #tan{(x-1)/(x+1)}={tantheta-tan(pi/4)}/{1+tantheta*tan(pi/4)},#

#=tan(theta-pi/4)#

#rArr y=arc tan{(x-1)/(x+1)}=arc tan{tan(theta-pi/4)}, i.e.,#

#y=theta-pi/4=arc tanx-pi/4, x in RR-{-1}.#

#:. dy/dx=1/(1+x^2)-0=1/(1+x^2), x in RR-{-1}.#

Enjoy Maths.!

Answer 2 · 2017-04-16T18:09:28

Recall that the derivative of #arctan(x)# is #1/(1+x^2)#. Then, by the chain rule, the derivative of #arctan(f(x))# is #1/(1+f(x)^2)*f'(x)#.

Here, this shows us that:

#d/dxarctan((x-1)/(x+1))=1/(1+((x-1)/(x+1))^2)d/dx((x-1)/(x+1))#

First simplifying the fraction:

#=(x+1)^2/((x+1)^2+(x-1)^2)d/dx((x-1)/(x+1))#

#=(x+1)^2/((x+1)^2+(x-1)^2)(((x+1)-(x-1))/(x+1)^2)#

#=2/((x+1)^2+(x-1)^2)#

#=2/((x^2+2x+1)+(x^2-2x+1))#

#=2/(2x^2+2)#

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