Question:
Y= sin-1(2x/1 x2) from inverse trigonometric functions of continuity and differentiability. Where has the root gone when you have solved the question in the video(Part 21)?
Answer:
Given, y = sin-1 {2x/(1 + x2 )}
We know that sin-1 x = 1/√(1 - x2 )
Now, differentiate w.r.t x, we get
dy/dx = d[sin-1 {2x/(1 + x2 )}]/dy
=> dy/dx = 1/√[1 - {2x/(1 + x2 )}2 ] * d{2x/(1 + x2 )}/dx
=> dy/dx = 1/√[1 - 4x2 /(1 + x2 )2 ] * d{2x/(1 + x2 )}/dx
=> dy/dx = (1 + x2 )/√[(1 + x2 )2 - 4x2 ] * [{d(2x/dx) * (1 + x2 ) - d(1 + x2 )/dx * 2x}/(1 + x2 )2 ]
=> dy/dx = (1 + x2 )/√[(1 + x2 )2 - 4x2 ] * [{2(1 + x2 ) - 2x * 2x}/(1 + x2 )2 ]
=> dy/dx = {(1 + x2 )/√(1 - x2 )2 } * [{2 + 2x2 - 4x2 }/(1 + x2 )2 ]
=> dy/dx = {(1 + x2 )/(1 - x2 )} * [{2 - 2x2 }/(1 + x2 )2 ]
=> dy/dx = {1/[(1 - x2 )} * [{2 - 2x2 }/(1 + x2 )]
=> dy/dx = {1/[(1 - x2 )} * [{2(1 - x2 )}/(1 + x2 )]
=> dy/dx = 2/(1 + x2 )
So, differentiation of sin-1 {2x/(1 + x2 )} is 2/(1 + x2 )
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