Question

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

Answer 1

Do you need the answer or explanation?

Answer 2

d?

Answer 3

It's for derivatives right?

Answer 4

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)

Answer 5

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.
\[dy/dx=dy/du.du/dx\]

Answer 6

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

Answer 7

\[d ^{2}y / dx ^{2}=-(-x ^{2}+25)^{-1/2}-x(-1/2)(-x ^{2}+25)^{-3/2}\]

Answer 8

i get
\[d^{2}y/dx = \frac{2x^{2} -25}{\sqrt{25-x^{2}}(25-x^{2})}\]

plug in x=4

Answer 9

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

Answer 10

is that the same as yours??
i'm off by a neg

Answer 11

uhhh, just the negative at the front of my equation is different

Answer 12

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