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anonymous

  • 4 years ago

If x^2+y^2=25, what is the value of d^2y/dx^2 at the point (4,3)?

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  1. anonymous
    • 4 years ago
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    Do you need the answer or explanation?

  2. anonymous
    • 4 years ago
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    d?

  3. anonymous
    • 4 years ago
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    It's for derivatives right?

  4. anonymous
    • 4 years ago
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    This may go wrong somewhere, but I think this is how it goes: x^2+y^2=25 y^2=-x^2+25 y=sqrt(-x^2+25)

  5. anonymous
    • 4 years ago
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    That's where we start. Now, we need to find the first derivative. We do this by using the formula for the derivatives of composite functions. \[dy/dx=dy/du.du/dx\]

  6. anonymous
    • 4 years ago
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    So, y=sqrt(u) u=(-x^2+25) y'=1/2(u^-1/2) (y' means derivative of y) u'=-2x Therefore, dy/dx=1/2(u^-1/2)(-2x) =-x(-x^2+25)^-1/2

  7. anonymous
    • 4 years ago
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    \[d ^{2}y / dx ^{2}=-(-x ^{2}+25)^{-1/2}-x(-1/2)(-x ^{2}+25)^{-3/2}\]

  8. dumbcow
    • 4 years ago
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    i get \[d^{2}y/dx = \frac{2x^{2} -25}{\sqrt{25-x^{2}}(25-x^{2})}\] plug in x=4

  9. anonymous
    • 4 years ago
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    Now, substituting 4 in, we should get the answer. -1/3+2/27 =-7/27

  10. dumbcow
    • 4 years ago
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    is that the same as yours?? i'm off by a neg

  11. anonymous
    • 4 years ago
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    uhhh, just the negative at the front of my equation is different

  12. dumbcow
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=second+derivative+y%3Dsqrt%2825-x%5E2%29+

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