Question

differentiate y=sin-¹x+sin-¹√1-x²​

Answer 1

Answer:

$differentiate y=sin-¹x+sin-¹√1-x²$

If y = sin⁻¹ x + sin⁻¹ √ (1 – x²), then how do you find dy/dx?

y=sin^-1(x)+sin^-1√(1-x^2).

Let sin^-1(x)=A. , then x= sinA

y= A+ sin^-1√(1-sin^2 A)

y= A + sin^-1(cosA).

or. y=A+sin^-1{sin(π/2-A)}.

or. y = A +π/2 -A

or. y= π/2.

or. dy/dx = 0. Answer.

Second -Method:-

y=sin^-1(x)+sin^-1√(1-x^2).

Formula. sin^-1(x)+sin^-1(y)=sin^-1{x.√(1-y^2)+y.√(1-x^2)}.

y=sin^-1{x.√(1–1+x^2)+√(1-x^2).√(1-x^2)}.

y = sin^-1{x.x+(1-x^2)}.

y= sin^-1(x^2+1-x^2)

y = sin ^-1(1)

y= π/2

dy/dx = 0. Answer.

Answer 2

Answer:

y=sin^-1(x)+sin^-1√(1-x^2).

Let sin^-1(x)=A. , then x= sinA

y= A+ sin^-1√(1-sin^2 A)

y= A + sin^-1(cosA).

or. y=A+sin^-1{sin(π/2-A)}.

or. y = A +π/2 -A

or. y= π/2.

or. dy/dx = 0. Answer.

Second -Method:-

y=sin^-1(x)+sin^-1√(1-x^2).

Formula. sin^-1(x)+sin^-1(y)=sin^-1{x.√(1-y^2)+y.√(1-x^2)}.

y=sin^-1{x.√(1–1+x^2)+√(1-x^2).√(1-x^2)}.

y = sin^-1{x.x+(1-x^2)}.

y= sin^-1(x^2+1-x^2)

y = sin ^-1(1)

y= π/2

dy/dx = 0. Answer.

Step-by-step explanation:

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