How do you differentiate $arctan (\frac{x - 1}{x + 1})$ $Arctan ((x-1)/(x+1))$ ?

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Answer 1 · 2017-04-14T13:16:56

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

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 · 2017-04-16T18:09:28

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:

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

$=(x+1)^2/((x+1)^2+(x-1)^2)(((x+1)-(x-1))/(x+1)^2)$

$=2/((x+1)^2+(x-1)^2)$

$=2/((x^2+2x+1)+(x^2-2x+1))$

$=2/(2x^2+2)$

$=1/(x^2+1)$

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