# Derivative Question

1. May 6, 2007

### mathmann

1. The problem statement, all variables and given/known data
For the Relation defined by x^5 + y^5 = 5 show that d2y/dx2.

2. Relevant equations

3. The attempt at a solution

x^5 + y^5 = 5
5x^4 + 5^4dy/dx = 0
d2y/dx2 = - 20x^3/20y^3

??

2. May 6, 2007

### Noober

$$x^5+y^5=5$$
$$5x^4dx+5y^4dy=0$$
$$5x^4dx=-5y^4dy$$
$$-\frac{5x^4}{5y^4}=\frac{dy}{dx}$$
$$\frac{5y^4(-20x^3)+5x^4(20y^3)}{25y^8}=\frac{d^2y}{dx^2}$$

$$\frac{d^2y}{dx^2}=\frac{100x^4y^3-100x^3y^4}{25y^8}$$

Last edited: May 6, 2007
3. May 7, 2007

### mathmann

The answer says d^2y/dx^2 = -20x^3/y^9.

??

4. May 8, 2007

### Kuno

d/dx(x^5) + d/dx(y^5) = d/dx(5)
5x^4 + 5y^4 * y' = 0
5x^4 = -5y^4 * y'
y' = 5x^4/(-5y^4)
y' = x^4/(-y^4)

y'' = [-y^4 * d/dx(x^4) - x^4 * d/dx(-y^4)]/(y^8)
y'' = [-4y^4 * x^3 + 4x^4y^3 * y']/(y^8)
substitute y'
y'' = [-4y^4 * x^3 + 4x^4 * y^3 * (x^4/(-y^4))]/(y^8)
y'' = [-4y^4 * x^3 - 4x^8/(y)]/(y^8)
multiply by y
y'' = -4[y^5 * x^3 + x^8)]/y^9
substitute y^5 = 5 - x^5
y'' = -4[(5 - x^5) * x^3 + x^8]/y^9
y'' = -4[5x^3 - x^8 + x^8]/y^9
y'' = -20x^3/y^9

Sorry, I haven't quite had time to try LaTeX yet....

5. May 8, 2007

### mathmann

Thank you very much Kuno.. great help.