derivative problem?

Question

what is the second derivative of x^3+y^3=28 at the point (3,1)?

y(3)=
i dont understand how to do this.

Answer 1

x^3+y^3=28
y^3 = 28 - x^3
y = (28 - x^3)^(1/3)

Use chain rule, by defining h(x) and g(x):
f(x) = (28 - x^3)^(1/3)
f(x) = h(g(x))
h(x) = x^1/3
g(x) = 28 - x^3
f'(x) = h'(g(x)) * (g'x)
f'(x) = 1/3(g(x))^-2/3 * -3x^2
f'(x) = 1/3(28 - x^3)^(-2/3) * -3x^2

Simplify by moving -3x^2 into the expression inside the ^2/3:
y = 1/3(28 - x^3)^(-2/3) * -3x^2
y = -(28 - x^3)^(-2/3) * x^2
y^(-3/2) = -(28 - x^3) * (x^2)^(-3/2)
y^(-3/2) = -(28 - x^3) * (x^-3)
y^(-3/2) = (1 - 28x^-3)
y = (1 - 28x^-3)^(-2/3)

So, now we have a simplified f'(x). We now derive again with the chain rule:
f'(x) = (1 - 28x^-3 - 1)^(-2/3)
f'(x) = h(g(x))
f''(x) = h'(g(x)) * g'(x)
g(x) = 1 - 28x^-3
h(x) = x^(-2/3)
h'(x) = -2/3x^(-5/3)
h'(g(x)) = -2/3(1 - 28x^-3)^(-5/3)
g'(x) = -3 * 28x^-4 = -84x^-4
f''(x) = -2/3(1 - 28x^-3)^(-5/3) * -84x^-4

Again, simplify by moving x^-4 into the expression inside the ^-5/3:
y = -2/3(1 - 28x^-3)^(-5/3) * -84x^-4
y = 56(1 - 28x^-3)^(-5/3) * x^-4
y^(-3/5) = (56)^(-3/5) * (1 - 28x^-3) * x^4^(-3/5)
y^(-3/5) = (56)^(-3/5) * (1 - 28x^-3) * x^-12/5)
y^(-3/5) = (56)^(-3/5) * (x^-12/5 - 28x^-27/5)
y = 56 (x^-12/5 - 28x^-27/5)^(-5/3)

So, our final answer is f''(x) = 56 (x^-12/5 - 28x^-27/5)^(-5/3)
f''(3) = 56 ((3)^-12/5 - 28(3)^-27/5)^(-5/3)
f''(3) = 56 (.071599 - .007425)
f''(3) = 56 (0.064174)
f''(3) = 3.593744 (solution!)

Answer 2

3x^2 + 3y^2(dy/dx) = 0

dy/dx = -(3x^2)/(3y^2)

That's the first derivative, good luck on finding the second derivative. Use the quotient rule to find the second derivative.