# Second derivative using implicit differentiation

1. Sep 29, 2011

### cphill29

1. The problem statement, all variables and given/known data

x^6 + y^6 = -6

I have to prove that y'' = 30x^4/y^11

2. Relevant equations

3. The attempt at a solution

Using implicit differentiation:

6x^5 + 6y^5 dy/dx = 0
6y^5 dy/dx = -6x^5
dy/dx = -x^5/y^5

Quotient Rule:

[(y^5)(-5x^4) - (-x^5)(5y^4 dy/dx)] / (y^5)^2
[-5x^4 y^5 + (x^5) 5y^4 dy/dx] / (y^5)^2

This is where I got stuck.

2. Sep 29, 2011

### CompuChip

You can replace dy/dx again, and then simplify a bit.
Also, you lost an equality sign somewhere... that expression on the last line, what is it equal to again?

3. Sep 29, 2011

### cepheid

Staff Emeritus
For me, rather than using the quotient rule, it was easier to do this:

$$y^5 \frac{dy}{dx} = -x^5$$

and differentiate this implicitly (using the product rule on the left-hand side). As CompuChip pointed out, you'll have to substitute in the expression for dy/dx, and you also get a handy substitution from the original statement that x^6 + y^6 = -6 .