Question

How can you differentiate `sin^-1((2x)/(1+x^2))` ?

Answer 1

`(dy)/(dx)=(2(1-x^2))/(|1-x^2|(1+x^2))`

Explanation:

Let ,

`y=sin^-1((2x)/(1+x^2))=arcsin((2x)/(1+x^2))`

Let ,

`y=arcsinu` , where , `color(green)(u=(2x)/(1+x^2)`

So, `(dy)/(du)=1/sqrt(1-u^2) and`

`(du)/(dx)=((1+x^2)*2-2x(0+2x))/(1+x^2)^2 to[because "quotient -rule"]`

`:.(du)/(dx)=(2+2x^2-4x^2)/(1+x^2)^2=(2(1-x^2))/(1+x^2)^2`

`"Using " color(blue) "Chain Rule :"`

`color(blue)((dy)/(dx)=(dy)/(du)(du)/(dx`

`:.(dy)/(dx)=1/sqrt(1-u^2)xx(2(1-x^2))/(1+x^2)^2`

Subst. `color(green)(u=(2x)/(1+x^2)` ,we get

`(dy)/(dx)=1/sqrt(1-((2x)/(1+x^2))^2 )xx(2(1-x^2))/(1+x^2)^2`

`=(1+x^2)/sqrt((1+x^2)^2-4x^2) xx(2(1-x^2))/(1+x^2)^2`

`=(1+x^2)/sqrt((1-x^2)^2) xx(2(1-x^2))/(1+x^2)^2tox!=1`

`=(1+x^2)/|1-x^2| xx(2(1-x^2))/(1+x^2)^2to[becausesqrt((a)^2)=|a|`

`=(2(1-x^2))/(|1-x^2|(1+x^2))`