nevermind, I got it.

[JohnCU]

y^2(y^2 - 4) = x^2(x^2 - 5)

Take the derivative of both sides...

(y^2)(2y) + (2yy')(y^2 - 4) = (x^2)(2x) + (2x)(x^2 - 5)
2y^3 + 2y^3y' - 8yy' = 4x^3 - 10

I think this is where I'm messing up, cause after I finish working this, I get a final answer but my Ti-89 says that it's the wrong answer...

Should be 2y^3y' twice. oops.

 

[RIGorous1]

Originally posted by: JohnCU
y^2(y^2 - 4) = x^2(x^2 - 5)

Take the derivative of both sides...

(y^2)(2y) + (2yy')(y^2 - 4) = (x^2)(2x) + (2x)(x^2 - 5)
2y^3 + 2y^3y' - 8yy' = 4x^3 - 10

I think this is where I'm messing up, cause after I finish working this, I get a final answer but my Ti-89 says that it's the wrong answer...

I would help you but I have no pencil with me... damn computers! If no one helps you in 5 mins I'll go look for one ... here's a tip for passing calculus: lose the 89. I've had plenty of friends rely on that thing so much they failed.

 

[CrazyPerson]

oh good... i was just going to explain it to you

 

[RIGorous1]

Originally posted by: JohnCU
y^2(y^2 - 4) = x^2(x^2 - 5)

Take the derivative of both sides...

(y^2)(2y) + (2yy')(y^2 - 4) = (x^2)(2x) + (2x)(x^2 - 5)
2y^3 + 2y^3y' - 8yy' = 4x^3 - 10

I think this is where I'm messing up, cause after I finish working this, I get a final answer but my Ti-89 says that it's the wrong answer...

Should be 2y^3y' twice. oops.

what happened to the y' after you took the product rule for the inside of (y^2-4) one the first one? the derivative should be (2yy') not just (2y)... damn the prime notation... with liebniz's notation you wouldn't make that mistake.