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

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Answer 1 · 2018-07-03T04:49:59

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

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 , $color(brown)(theta=arc cotx$

$y=-1/2*arc cotx$

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

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

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