Asked by Millie
If x^2=25-y^2, what is the value of [(d^2)(y)]\[dx^2] at the point (3,4)?
Answers
Answered by
Reiny
first derivative:
2x = -2y dy/dx
at (3,4)
dy/dx = -3/4
2nd derivative:
2 = -2y( d^2y/d^2x) + (-2)(dy/dx)
at the given point (3,4)
2 = -8(d^2 y/d^2 x) - 2(-3/4)
8 (d^2 y/d^2 x ) = 3/2 - 2 = -1/2
d^2 y/d^2 x = -1/16
2x = -2y dy/dx
at (3,4)
dy/dx = -3/4
2nd derivative:
2 = -2y( d^2y/d^2x) + (-2)(dy/dx)
at the given point (3,4)
2 = -8(d^2 y/d^2 x) - 2(-3/4)
8 (d^2 y/d^2 x ) = 3/2 - 2 = -1/2
d^2 y/d^2 x = -1/16
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