Question

Question 12: Find dy/dx : y = sin¯¹(1-x²/ 1+x²), 0 < x < 1

Class 12 - Math - Continuity and Differentiability

Answer 1

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

$y = {sin}^{ - 1} ( \frac{1 - {x}^{2} }{1 + {x}^{2} } ) \\ \\ put \: \: x = tan \theta \\ \\ y = {sin}^{ - 1} ( \frac{1 - {tan}^{2} \theta }{1 + {tan}^{2} \theta } ) \\ \\ y = {sin}^{ - 1} (cos2 \theta) \\ \\ y= {sin}^{ - 1} (sin( \frac{\pi}{2} - 2 \theta)) \\ \\ y = \frac{\pi}{2} - 2\theta \\ \\ y = \frac{\pi}{2} - 2{tan}^{ - 1} x \\ \\ \frac{dy}{dx} = \frac{ - 2}{1 + {x}^{2} }$