Question

How do you find the derivative of Inverse trig function `y= arctan(x - sqrt(1+x^2))`?

Answer 1

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

Explanation:

Here,

`y=arctan(x-sqrt(1+x^2))`

We take , `x=cottheta , where, theta in (0,pi)`

`=>theta=arc cot x, x in RR and theta/2 in (0.pi/2)`

`=>y=arctan(cottheta-sqrt(1+cot^2theta))`

`=>y=arctan(cottheta-csctheta)...toapply(1)`

`=>y=arctan[-(csctheta-cottheta)]`

`=>y=-arctan(csctheta-cottheta)...toApply(6)`

`=>y=-arctan(1/sintheta-costheta/sintheta)`

`=>y=-arctan((1-costheta)/sintheta)...toApply(2) and (3)`

`=>y=-arctan((2sin^2(theta/2))/(2sin(theta/2)cos(theta/2)))`

`=>y=-arctan(sin(theta/2)/cos(theta/2))`

`=>y=-arctan(tan(theta/2)) ,where, theta/2 in(0,pi/2)`

`=>y=-theta/2...to Apply(4)`

Subst, back , `theta=arc cot theta`

`y=-1/2arc cot theta`

`=>(dy)/(dx)=-1/2(-1/(1+x^2))...toApply(5)`

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

Note: (Formulas)

`(1)1+cot^2theta=csc^2theta`

`(2)1-costheta=2sin^2(theta/2)`

`(3)sintheta=2sin(theta/2)cos(theta/2)`

`(4)arctan(tanx)=x`

`(5)d/(dx)(arc cot x)=-1/(1+x^2)`

`(6)arctan(-x)=-arctanx`