Finding second derivative at (3,2)?

x^2 + y^2 = 25. 2x + 2yy' = 0, ie., x + yy' = 0...(1). 1 + yy'' +(y')^2 = 0...(2).

At (2,3), (1)--->2 + 3y' = 0, ie., y' = -(2/3) and (2)--->1 + 3y'' + (-2/3)^2 = 0, ie., 

y'' = -(1/3)(1 + 4/9) = -(1/3)(13/9) = - 13/27.

x² + y² = 25 → when: x = 3

9 + y² = 25

y² = 16

y = ± 4

→ Point A (3 ; 4)

→ Point B (3 ; - 4)

x² + y² = 25 → when x is constant: → 2y.dy = 0

x² + y² = 25 → when y is constant: → 2x.dx = 0

2y.dy + 2x.dx = 0

2y.dy = - 2x.dx

dy/dx = - 2x/2y

dy/dx = - x/y

(dy/dx)² = - (x/y)²

d²y/d²x = - x²/y²

d²y/d²x = - x²/y² → @ the point A (3 ; 4)

d²y/d²x = - 9/16

d²y/d²x = - x²/y² → @ the point B (3 ; - 4)

d²y/d²x = - 9/16

(3,2) is NOT on the curve !

x² + y² = 25

Differentiation with respect to x

2x + 2y dy/dx = 0

dy/dx = -2x/2y

=-x/y at point (2, 3)

=-2/3

Again differentiation with respect to x

d²x/dy² = 1(y) – xy’/y² at (2, 3)

d²x/dy² = 3/(3)² - 2(- 2/3)/(3)²

= 1/3 + 4/27

= 13/27

If (a,b) is a point on x²+y²=r², so a²+b²=r², then:

1) dy/dx at (a,b) is the slope of the tangent line at that point. The line has the equation ax+by=r², so dy/dx = -a/b and so dx/dy = -b/a.

2) d²y/dx² at (a,b) is the slope of the radial line at that point. The line has the equation bx-ay=0, so d²y/dx² = b/a, and so d²x/dy² = a/b.

Here d²x/dy² at (3,2) would equal 3/2 for x²+y²=13.

Your update gets us no closer to an answer. Neither (3, 2) nor (2, 3) satisfies the equation you gave. I am not saying that the answer cannot be found. What I mean is that the answer cannot even exist.

2x dx + 2y dy = 0 =>

dy/dx = -x/y =>

d2y/dx2 = (-y + x dy/dx)/y^2.

Did you mean (3,4) ? The point (3,2) is not on the given circle!

Anyway, to find d2y/dx2 at some point that IS on the circle, just plug the numbers into the above formula, first calculating dy/dx.