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

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

Mathematics
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d?
It's for derivatives right?

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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)
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. \[dy/dx=dy/du.du/dx\]
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
\[d ^{2}y / dx ^{2}=-(-x ^{2}+25)^{-1/2}-x(-1/2)(-x ^{2}+25)^{-3/2}\]
i get \[d^{2}y/dx = \frac{2x^{2} -25}{\sqrt{25-x^{2}}(25-x^{2})}\] plug in x=4
Now, substituting 4 in, we should get the answer. -1/3+2/27 =-7/27
is that the same as yours?? i'm off by a neg
uhhh, just the negative at the front of my equation is different
http://www.wolframalpha.com/input/?i=second+derivative+y%3Dsqrt%2825-x%5E2%29+

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