How do you differentiate #Arctan ((x-1)/(x+1))#?
↳Redirected from
"What is the freezing point depression?"
# 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.!
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))#
Using the quotient rule:
#=(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)#
