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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)