Question

​find derivative of the above question with respect to x

Answer 1

Answer

dy/dx = √( 1 - y² ) / √ ( 1 - x² )

Step-by-step explanation:

Given --->

√( 1 - x² ) + √( 1 - y² ) = a ( x - y )

Concept used --->

1) First we used substitution as

x = Sin α and y = Sinβ

2) Then we solve it and differentiate by using formulee of differentition

a ) d / dx ( sin⁻¹ x ) = 1 / √( 1 - x² )

Solution -----> ATQ,

√( 1 - x² ) + √( 1 - y² ) = a ( x - y )

Let

x = Sinα , y = Sinβ

=> α = Sin⁻¹ x , β = Sin⁻¹ ( y )

=> √( 1 - Sin²α ) + √( 1 - Sin²β ) = a ( Sinα - Sinβ )

We know that , 1 - Sin²θ = Cos²θ , applying it we get,

=> √( Cos²α ) + √( Cos²β ) = a ( Sinα - Sinβ )

=> Cosα + Cosβ = a ( Sinα - Sinβ )

Now , we have two formulee , as follows ,

CosC + CosD = 2 Cos ( C + D / 2 ) Cos ( C - D / 2 )

SinC - SinD = 2 Cos ( C + D / 2 ) Sin ( C - D / 2 )

Applying these we get,

=> 2 Cos(α + β / 2 ) Cos ( α - β/2 )

= a 2 Cos ( α + β /2 ) Sin ( α - β / 2 )

2 Cos ( α + β / 2 ) is cancel out from each side ,

=> Cos ( α - β / 2 ) = a Sin ( α - β / 2 )

=> Cos ( α - β / 2 ) / Sin ( α - β / 2 ) = a

=> Cot ( α - β / 2 ) = a

=> Cot⁻¹ Cot ( α - β / 2 ) = Cot⁻¹ a

=> ( α - β ) / 2 = Cot⁻¹ a

=> α - β = 2 Cot⁻¹ a

=> Sin⁻¹ x - Sin⁻¹ y = 2 Cot⁻¹ a

Now , differentiating with respect to x , we get ,

=> d/dx ( Sin⁻¹x ) - d/dx ( Sin⁻¹y ) = d/dx ( 2Cot⁻¹a )

=> 1/√(1 - x²) - 1 / √( 1 - y² ) dy/dx = 0

=> 1 / √( 1 - y² ) dy/dx = 1 / √( 1 - x² )

=> dy/dx = √( 1 - y² ) / √( 1 - x² )