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

Question

8696 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-05T17:49:01

# -1/(2sqrt(1-x^2)), -1 lt x lt 1#.

Let, #y=arc tansqrt((1-x)/(1+x)#.

Now, note that, #sqrt((1-x)/(1+x)# will be meaningful, iff,

# (1-x) ge 0, and (1+x) gt 0, i.e., -1 lt x le 1#.

Also, the Range of #cos# function is #[-1,1]#.

We can, hence, subst., #x=costheta," so that, "theta=arc cos x#.

Then, we have, #y=arc tansqrt((1-costheta)/(1+costheta))#,

#=arc tansqrt((2sin^2(theta/2))/(2cos^2(theta/2))#,

#=arc tan(tan(theta/2))#,

#=theta/2#.

#:. y=1/2arc cosx#,

#:. dy/dx=1/2*-1/sqrt(1-x^2)=-1/(2sqrt(1-x^2)), -1 lt x lt 1#.

Enjoy Maths.!