Question

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

Answer 1

`-1`

Explanation:

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`

Answer 2

`"The Reqd. Deri.="1, if -1ltxlt1;`
`=-1, if x>1:`
`-1, if x<-1.`

Explanation:

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.`