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Question 726751: I have a question regarding finding the second derivative of a hyperbola...
Find the value of d^2y/dx^2 for the hyperbola defined by the equation y^2-x^2=7 at the point (3,4).
The first derivative I have already found- dy/dx=x/y
However, to find the second derivative, I was a little bit more confused...
y=xy^-1
d/dx(y)=d/dx(xy^-1)
dy/dx=x*dy/dx*-1y^-2 + y^-1 * d/dx (x)
dy/dx= -x*dy/dx*y^-2 +y^-1
dy/dx- (dy/dx (-xy^-2))= y^-1
dy/dx(1+xy^-2)= y^-1
dy/dx= y^-1/1+xy^-2
dy/dx= 1/y(1+xy^-2)
Now, to plug in the points (3,4)
dy/dx= 1/y(1+xy^-2)
1/(4(1+(3)(4^-2))
= 1/(4+ (12/6)
I hope this was not too confusing- but I would like to know if this is the correct method for solving this problem, and if the answer is correct as well.
I would really appreciate your help.
Thank you very much.
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Find the value of d^2y/dx^2 for the hyperbola defined by the equation y^2-x^2=7 at the point (3,4).
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2ydy - 2xdx = 0
y' = x/y (just checking)
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y' = xy^-1
y'' = y^-1*dx + x*(-1)*y^-2*dy
The first derivative I have already found- dy/dx=x/y
However, to find the second derivative, I was a little bit more confused...
y=xy^-1
d/dx(y)=d/dx(xy^-1)
dy/dx=x*dy/dx*-1y^-2 + y^-1 * d/dx (x)
dy/dx= -x*dy/dx*y^-2 +y^-1
dy/dx- (dy/dx (-xy^-2))= y^-1
dy/dx(1+xy^-2)= y^-1
dy/dx= y^-1/1+xy^-2
dy/dx= 1/y(1+xy^-2)
============================
(3/4) is in the 1st Q, so solve for the + half of the hyperbola
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y = (x^2 + 7)^1/2
y' = (1/2)*(x^2+7)^(-1/2)*2x
y' = x*(x^2+7)^(-1/2)
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y'' = (x^2+7)^(-1/2) + x*(-1/2)*(x^2+7)^(-3/2)*2x
y'' = ((x^2+7)^(-1/2) - (x^2))/(x^2+7)^(3/2)
y'' = (x^2+7 - x^2)/((x^2+7)^(3/2))
y'' = 7/(x^2+7)^3/2)
y''(3) = 7/(16)^3/2
= 7/64
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