# What is the derivative of arctan(x - sqrt(1+x^2))?

Jul 3, 2018

=>(dy)/(dx)=1/(2(1+x^2)

#### Explanation:

We know that,

color(blue)((1)1-costheta=2sin^2(theta/2) and sintheta=2sin(theta/2)cos(theta/2)

color(violet)((2)arc tan (-alpha)=-arc tanalpha

color(green)((3)arc tan(tanphi)=phi ,where,phi in(-pi/2,pi/2)

Let

$y = a r c \tan \left(x - \sqrt{1 + {x}^{2}}\right)$

We take, color(brown)(x=cottheta=>theta=arc cotx,where,theta in(0,pi)=>theta/2in(0,pi/2)

So,

y=arc tan(cottheta-sqrt(1+cot^2theta))to[use : color(red)(1+cot^2theta=csc^2theta]

$\implies y = a r c \tan \left(\cos \frac{\theta}{\sin} \theta - \csc \theta\right)$

$\implies y = a r c \tan \left(\cos \frac{\theta}{\sin} \theta - \frac{1}{\sin} \theta\right)$

=>y=arc tan((costheta-1)/sintheta)...tocolor(violet)(Apply(2)

$\implies y = a r c \tan \left\{- \left(\frac{1 - \cos \theta}{\sin} \theta\right)\right\}$

=>y=-arc tan((1-costheta)/sintheta)...tocolor(blue)(Apply(1)

$\implies y = - a r c \tan \left(\frac{2 {\sin}^{2} \left(\frac{\theta}{2}\right)}{2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}\right)$

$\implies y = - a r c \tan \left(\frac{\sin \left(\frac{\theta}{2}\right)}{\cos} \left(\frac{\theta}{2}\right)\right)$

=>y=-arc tan(tan(theta/2)).tocolor(green)(Apply(3) ,as ,theta/2in(0,pi/2)

$\implies y = - \frac{\theta}{2}$

Subst. back , color(brown)(theta=arc cotx

$y = - \frac{1}{2} \cdot a r c \cot x$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2} \cdot \left(- \frac{1}{1 + {x}^{2}}\right)$

=>(dy)/(dx)=1/(2(1+x^2)