Question #efa3a

1 Answer
Jan 13, 2018

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

Explanation:

.

#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)#