Asked by Gwen
If y = (x + squareroot x power2 + 1)power 2 show that dy over dx = 2y over x square root x power 2 +1
Answers
Answered by
oobleck
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)
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)
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