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

Question

Paisley Johnson · Accepted Answer

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

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=arc tan(x-sqrt(1+x^2))#

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]#

#=>y=arc tan(costheta/sintheta-csctheta)#

#=>y=arc tan (costheta/sintheta-1/sintheta)#

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

#=>y=arc tan{-((1-costheta)/sintheta)}#

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

#=>y=-arc tan((2sin^2(theta/2))/(2sin(theta/2)cos(theta/2)))#

#=>y=-arc tan((sin(theta/2))/cos(theta/2))#

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

#=>y=-theta/2#

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

#y=-1/2*arc cotx#

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

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

Evelyn Agard · Answer

undefined The derivative of $ \arctan(x - \sqrt{1+x^2}) $ is $ \frac{1}{1+(x - \sqrt{1+x^2})^2}\left(1-\frac{x}{\sqrt{1+x^2}}\right) $.

Isabella Ackerman · Answer

undefined The derivative of $ \arctan(x - \sqrt{1+x^2}) $ is $ \frac{-1}{\sqrt{1+x^2} \cdot \sqrt{2 - 2x - 2\sqrt{1+x^2}}} $.