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

Question

its the second derivative but im not sure how to get there.

Answer 1

Just use implicit differentiation to find the derivatives:
2x+2y*y'=0, solve for y'
y'=-x/y Now you can find the second derivative using the product rule or rewriting it to use the multiplication rule. I like the multiplication rule better- it's easier to remember.
y'=-x*y^-1
y''=-y^-1+(-1)*-x*y^-2, now simplify
y''=(-1/y)+(x/y^2) and in its "prettiest" from it looks like this:
y''=(x-y)/y^2

Now, just substitute in the point (4,3)
y''=(4-3)/(3^2)
y''=1/9

Answer 2

Differentiate implicitly...
2x + 2yy' = 0.

At (4,3) this becomes...
2(4) + 2(3)y' = 0
→ y' = -4/3.

Differentiate again...
2 + 2yy" + 2(y')² = 0
→ yy" = - [1 + (y')²]
→ y" = - [1 + (y')²] / y
At (4,3) this becomes y" = - [1 + (-4/3)²] / 3 = -25/27.

So... three different answers.

I may have made a mistake in calculation, but at least my answer is negative... as it should be. Draw the circle and you'll see that it's concave down at (4,3).

P.S. I spotted the mistakes above me... in both cases, fixing them makes the answer -25/27.

Quantum's error:
y'=-x*y^-1
y''=-y^-1+(-1)*-x*y^-2,
the last term should be (-1)*-x*y' *y^-2 (chain rule says you need y' in there).

iyogrenci says the derivative of 2x+2yy'=0 is
2+2(y' y' +y'' y' )=0... this should be 2+2(y' y' +y'' y )=0
(y intead of y' in the last term between brackets) if you apply the product rule properly.

Answer 3

the first derivative
2x+2yy'=0

x=4
y=3

8+6y'=0

y' = -4/3


the second derivative
2+2(y' y' +y''y')=0
2+2(16/9+y''(-4/3)=0

8/3y''=50/9

y''=25/12