Question

Y = sin-¹(1)/(√1+x²)

Answer 1

Answer:

Y = sin¯¹(1-x²/ 1+x²)

put x = tan∅

y = sin-¹(1-tan²∅/1+tan²∅)

y = sin-¹(cos2∅) { 1-tan²∅/1+tan²∅ = cos2∅

y = sin-¹{sin(π/2-2∅)}

Now 0<x<1 => 0<tan∅ <1 => 0<∅<π/4

=> 0<π/2-2∅<π/2

from this

y= π/2-2∅

differentiating w.r.t x

y = π/2-2tan-¹x

dy/dx = -2/1+x² { d/dx(tan-¹x = 1/1+x2

Answer 2

Answer:

I think it's y = sin-¹ x/(√1+x²) , then the Answer is dy/dx = (1-xy) / (1+x²)

Step-by-step explanation:

y = sin-¹ x/(√1+x²)

differentiating both the sides with respect to X

dy/dx = [√1+x² (1/√1+x²) - sin-¹ x {1/2√1+x²}]/ (√1+x²)²

dy/dx = {1-(x sin -¹x/√1+x²)}/(√1+x²)²

(1+x²) dy/dx = 1- x . y

{ Because , y = sin-¹ x/(√1+x²) }

dy/dx = (1-xy) / (1+x²)

Hope it will be helpful to You....!!!!