# Implicit Derivative

1. Nov 14, 2004

Hi all:

How would you find the second derivative of

2x^3 - 3y^2 = 8?

I know the first derivative is x^2 / y. Would I use the quotient rule, or would I use some type of substitution and then use the product rule?

Any help is greatly appreciated!

Thanks

2. Nov 14, 2004

### nrqed

Maybe I am wrong because it sounds too easy (so I might be forgetting something), but isn't it symply given by differentiating dy/dx a second time, which will give

$${d^2 y\over dx^2} = {d \over dx} {x^2 \over y} = {2 x \over y} - {x^2 \over y^2} {dy \over dx} = {2 x \over y} -{ x^4 \over y^3}$$

I might be missing something because it looked too simple!

Pat

3. Nov 14, 2004

### Hurkyl

Staff Emeritus
You're missing a bit of specification here... I'm guessing from your partial result you're looking for the second derivative of y with respect to x.

When I'm unsure about a question, I often do two things. Here, I would:

(a) Try to come up with reasons why I couldn't use the quotient rule.
(b) Look for justification for using the quotient rule.

Have you gotten anywhere on either of these?

(edit: *sigh* there goes my attempt at a confidence building exercise)

4. Nov 14, 2004

ok Thanks a lot

5. Nov 14, 2004

### nrqed

I am sorry, really.

Pat

6. Nov 15, 2004

### Hurkyl

Staff Emeritus
(don't be sorry)

7. Nov 15, 2004

### HallsofIvy

Staff Emeritus
2x3- 3y2= 8 so
6x2- 6y y'= 0.

Differentiate again:

12x- (6y')y'- 6yy"= 0 or

12x- 6y'2- 6yy"=0.

Solve for y".