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

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Answer 1 · 2018-06-09T18:36:21

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

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#