Question

How do you differentiate `Arctan ((x-1)/(x+1))`?

Answer 1

`1/(1+x^2), x!=-1, x in RR.`

Explanation:

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.!

Answer 2

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)`