View Full Version : Implicit Derivative
courtrigrad
Nov14-04, 07:58 PM
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
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
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
Nov14-04, 08:08 PM
ok Thanks a lot
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:
HallsofIvy
Nov15-04, 06:53 AM
2x3- 3y2= 8 so
6x2- 6y y'= 0.
Differentiate again:
12x- (6y')y'- 6yy"= 0 or
12x- 6y'2- 6yy"=0.
Solve for y".
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