Asked by Anonymous
If y^2=1+(2/x) find (d^2y)/dx^2 in terms of x and y
Answers
Answered by
Reiny
y^2 = 1 + 2/x
2y dy/dx = -2/x^2
y' = -2/(2yx^2) = -1/(y x^2)
using the quotient rule,
y '' = (yx^2(0) - (-1)(x(2x) + x^2 dy/dx) )/(y^2 x^4)
= (2xy + x^2(-1/(yx^2) )/(y^2x^4)
= (2xy - 1/y)/(y^2 x^4)
= (2x y^2 - 1)/(y^3 x^4)
check my algebra
2y dy/dx = -2/x^2
y' = -2/(2yx^2) = -1/(y x^2)
using the quotient rule,
y '' = (yx^2(0) - (-1)(x(2x) + x^2 dy/dx) )/(y^2 x^4)
= (2xy + x^2(-1/(yx^2) )/(y^2x^4)
= (2xy - 1/y)/(y^2 x^4)
= (2x y^2 - 1)/(y^3 x^4)
check my algebra