Implicit Derivative

[courtrigrad]

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

[nrqed]

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

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

[Hurkyl]

How would you find the second derivative

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)

[courtrigrad]

ok Thanks a lot

[nrqed]

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)

I am sorry, really.

Pat :frown:

[Hurkyl]

(don't be sorry)

[HallsofIvy]

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

Differentiate again:

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

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

Solve for y".