Question #efa3a

1 Answer
Jan 13, 2018

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

Explanation:

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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)