y=sin^(-1) [(1+x)^1/2+(1-x)^1/2]/2

Question

You may check your answer Answer for above question is ..... Dy/dx= -1/2(a^2-x^2)^1/2

Answer 1

When y=sin^-1(x/a), sin(y)=x/a, cos(y)=√(1-x²/a²), cos(y)dy/dx=1/a, √(1-x²/a²)dy/dx=1/a, dy/dx=1/√(a²-x²).

But this problem has:

sin(y)=½(√(1+x)+√(1-x)),

sin²(y)=¼(1+x+1-x+2√(1-x²))=½(1+√(1-x²)),

cos(y)=√(1-sin²(y))=√(1-½(1+√(1-x²))=√(½(1-√(1-x²))),

cos(y)dy/dx=¼(1/√(1+x)-1/√(1-x))=¼(√(1-x)-√(1+x))/√(1-x²),

√(½(1-√(1-x²)))dy/dx=¼(√(1-x)-√(1+x))/√(1-x²),

dy/dx=¼(√(1-x)-√(1+x))/[√(1-x²)√(½(1-√(1-x²)))].