If y = (x + squareroot x power2 + 1)power 2 show that dy over dx = 2y over x square root x power 2 +1

Answer 1

using more conventional math notation,
y = (x + √(x^2+1))^2
y' = 2(x + √(x^2+1))(1 + 1/(2√(x^2+1))*2x)
= 2(x + √(x^2+1))(x + √(x^2+1))/√(x^2+1)
= 2(x + √(x^2+1))^2/√(x^2+1)
= 2y/√(x^2+1)