Differentiate #tan^-1 ((2x)/(1-x^2))# with respect to #sin^-1 ((2x)/(1+x^2))#?

Calling

#{(y_1=arctan(f_1(x))),(y_2=arcsin(f_2(x))):}#

with

#{(f_1(x)=(2x)/(1-x^2)),(f_2(x)=(2x)/(1+x^2)):}#

we need #(dy_1)/(dy_2)#

but

#(dy_1)/(dy_2)=((dy_1)/(dx))//((dy_2)/(dx)) = (dy_1)/(dx) (dx)/(dy_2) = (dy_1)/(dy_2)#

and

#{(d/(dx)arctan(f_1(x))=(f'_1(x))/(1+f_1^2(x))),(d/(dx)arcsin(f_2(x))=(f'_2(x))/sqrt(1-f_2^2(x))):}#

note that

#d/(dx)arctan(g(x))=d/(dg)arctan(g)(dg)/(dx)# and
#d/(dx)arcsin(g(x))=d/(dg)arcsin(g)(dg)/(dx)#

and

#{(d/(dg)arctan(g)=1/(1+g^2)),(d/(dg)arcsin(g)=1/sqrt(1-g^2)):}#

so after substituting

#{(f'_1(x)=(2 (1 + x^2))/(x^2-1)^2),(f'_2(x)=(2 (1 - x^2))/(1 + x^2)^2):}#

and simplifying

#(dy_1)/(dy_2)=((f'_1(x))/(1+f_1^2(x)))/((f'_2(x))/sqrt(1-f_2^2(x)))=-1#

Let #u=tan^-1((2x)/(1-x^2)), and, v=sin^-1((2x)/(1+x^2)).#

Note that, because of the Dr. of #u, x in RR-{+-1}....(ast)#

#:. x <-1, or, -1 lt x lt 1, or, x >1.#

Subst. #x=tantheta. because (ast), theta in (-pi/2,pi/2)-{+-pi/4}, &, theta=tan^-1x.#

#:. u=tan^-1{(2tantheta)/(1-tan^2theta)}=tan^-1(tan2theta), and, #

#v=sin^-1{(2tantheta)/(1+tan^2theta)}=sin^-1(sin2theta).#

Case (1) : #-1 lt x lt 0, and, 0 lt x lt 1.#

#:. tan(-pi/4) lt tantheta lt tan 0, &, tan0 lt tantheta lt tan(pi/4).#

Since, #tan# fun. is #uarr# in all quadrants, it follows that,

#-pi/4 lt theta lt 0, and, 0 lt theta lt pi/4.#

#:. -pi/2 lt 2theta lt 0, &, 0 lt 2theta lt pi/2.#

&, #:.# by the Defns of #tan^-1 and sin^-1# functions, we have,

#u=tan^-1(tan2theta)=2theta, and, v=sin^-1(sin2theta)=2theta.#

Thus, #u=2tan^-1x=v, if, -1 lt x lt 1.#

#"Therefore, the Reqd. Deri.="(du)/(dv)={(du)/dx}/{(dv)/dx},#

#={2/(1+x^2)}/{2/(1+x^2)}=1, if -1 lt x lt 1.#

Case (2) : #x > 1.#

#:. tantheta > tan(pi/4) rArr theta > pi/4...[because, tan" is "uarr"]#

Preferably, #pi/4 < theta < pi/2 rArr pi/2 < 2theta < pi.#

#rArr pi/2-pi < 2theta-pi < pi-pi, i.e., -pi/2 <2theta-pi < 0.#

Then, #tan(2theta -pi)=-tan(pi-2theta)=-(-tan2theta)=tan2theta.#

#:. u=tan^-1(tan2theta)=tan^-1(tan(2theta-pi))," where "(2theta-pi) in (-pi/2,0) sub (-pi/2,pi/2).#

#:.# by the Defns. of #tan^-1 and sin^-1# functions, we get,

#u=2theta-pi=2tan^-1x-pi;#

#"Also, "sin(2theta-pi)=-sin(pi-2theta)=-sin2theta#

#:. sin2theta=-sin(2theta-pi).#

#:. v=sin^-1(sin2theta)=sin^-1(-sin(2theta-pi))=-sin^-1(sin(2theta-pi))=-(2theta-pi)=pi-2tantheta=pi-2tan^-1x, (x >1)#

#:." The Reqd. Deri.="{2/(1+x^2)-0}/{0-2/(1+x^2)}=-1, if x >1.#

Case (3) : #x lt -1.#

In this case, #x lt -1 rArr theta lt -pi/4.#

We take, #-pi/2 lt theta lt -pi/4 :. -pi lt 2theta lt -pi/2.#

#:. 0 lt (pi+2theta) lt pi/2 rArr (pi+2theta) in (0,pi/2) sub (-pi/2,pi/2).#

Also, #tan(pi+2theta)=tan2theta, &, sin(pi+2theta)=-sin2theta.#

#:. u=tan^-1(tan2theta)=tan^-1(tan(pi+2theta))=pi+2theta=pi+2tan^-1x,#

and, #v=sin^-1(sin2theta)=sin^-1(-sin(pi+2theta))=-sin^-1(sin(pi+2theta))=-pi-2theta=-pi-2tan^-1x, (xlt-1.)#

#:. (du)/(dv)=-1, xlt-1.#