How do you differentiate #\tan ^ { - 1} ( \cos \sqrt { x } )#?

The derivative of #arctanu# is #1/(1 + u^2)(du)/(dx)#. If you do not remember that, I have derived it at the bottom.

I got

#-(sinsqrtx)/(2sqrtx(1 + (cos sqrtx)^2))#

Lots of chain rule going on here. If we had #f(u) = arctan(u)#, and #u(v) = cos(v)#, then #v = sqrtx#, and:

#(df)/(dx) = (df)/(du) (du)/(dv) (dv)/(dx)#

#= d/(du)[f(u)] cdot d/(dv)[u(v)] cdot d/(dx)[v(x)]#

#= d/(du)[arctanu] cdot d/(dv)[cosv] cdot d/(dx)[sqrtx]#

#= 1/(1 + u^2) cdot (-sinv) cdot 1/(2sqrtx)#

Therefore, since #u = cossqrtx# and #v = sqrtx#:

#d/(dx)[arctan(cossqrtx)]#

#= 1/(1 + (cossqrtx)^2) cdot (d(cossqrtx))/(dx) cdot (d(sqrtx))/(dx)#

#= 1/(1 + (cossqrtx)^2) cdot -sinsqrtx cdot 1/(2sqrtx)#

#= color(blue)(-(sinsqrtx)/(2sqrtx(1 + (cos sqrtx)^2)))#

Define #x = tany# so that #arctanx = y#. Then, using the chain rule,

#1 = sec^2y (dy)/(dx)#

As a result:

#(dy)/(dx) = 1/(sec^2y)#

Since #sec^2 y = 1 + tan^2y#:

#(dy)/(dx) = 1/(1 + tan^2y)#

Since #tany = x#, #tan^2y = x^2#, so with #y = arctanx#:

#color(green)((dy)/(dx) = d/(dx)[arctanx] = 1/(1 + x^2))#