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#arctanz# means an angle whose tangent is #z#. As such, if we name our angle #theta# it can be written as:
#tantheta=z#
Let's let #u=((x-1)/(x+1))# then:
#f(x)=arctanu#
which means an angle whose tangent is #u#. Let's call our angle #alpha#. Then we can rewrite the equation as:
#tanalpha=u# which is the same as:
#alpha=arctanu=arctan((x-1)/(x+1))#, #color(red)(Equation 1)#
Now, let's take derivatives of both sides of:
#tanalpha=u#
#sec^2alphadalpha=du#
#(dalpha)/(du)=1/sec^2alpha#
Using the identity #sec^2theta=1+tan^2theta#, we get:
#(dalpha)/(du)=1/(1+tan^2alpha)=1/(1+u^2)#
Earlier, we said:
#u=((x-1)/(x+1))#
Let's take derivatives of both sides using the Quotient Rule for the right hand side:
#du=(((x+1)-(x-1))/(x+1)^2)dx#
#(du)/dx=2/(x+1)^2#
The Chain Rule states:
#(dalpha)/dx=(dalpha)/(du)*(du)/dx#
Let's plug them in:
#(dalpha)/dx=1/(1+u^2)*2/(x+1)^2=1/(1+((x-1)/(x+1))^2)(2/(x+1)^2)#
#(dalpha)/dx=1/(((x+1)^2+(x-1)^2)/(x+1)^2)(2/(x+1)^2)#
#(dalpha)/dx=2/((x+1)^2+(x-1)^2)=2/(2(x^2+1))=1/(x^2+1)#
From Equation 1 above,
#d/dx(arctan((x-1)/(x+1)))=1/(x^2+1)#
