# Thread: someone wanna look over my work?

1. ## someone wanna look over my work?

i would greatly apperciate it if someone were to take a look at my work and tell where im wrong or right thx

oh yeah i'll probably be bugging ya guys all night

2. Originally Posted by cyberdx16
i would greatly apperciate it if someone were to take a look at my work and tell where im wrong or right thx

oh yeah i'll probably be bugging ya guys all night
First problem: Just a typo. Should be dx/dt = 3y^2/(6y). You had "x^2."

Third problem:
You have y' = 4/5 + (3/5)y'
So (5/5)y' - (3/5)y' = 4/5
(2/5)y' = 4/5
You've got an extra negative sign.

Fourth problem:
x^2 + y^3 = 5y

Take the first derivative:
2x + 3y^2*y' = 5 y'

Take the second derivative:
2 + 6y*y' + 3y^2*y'' = 5y''

3y^2*y'' - 5y'' = -2 - 6y*y'

y'' = -(6y*y' + 2)/(3y^2 - 5)

You've got the point (x,y) from the problem and y' at this point from the last problem.

And for the record, you can TRY to bug me all night, but I'm going to be in bed by 11.

-Dan

3. how did u get (the bold part?) y"=2+6y*y'+3y^2*y"=5y"

4. Originally Posted by cyberdx16
how did u get (the bold part?) y"=2+6y*y'+3y^2*y"=5y"
The product rule

Before the man differenciated he had,
3y^2*y'

Use the product rule

[3y^2]'y'+[3y^2][y']'
Thus,
[6yy'][y']+[3y^2][y'']
Thus,
6y(y')^2+3y^2y'' (He made an error)

5. so does anyone know if i did the 2nd problem right?

6. Originally Posted by cyberdx16
so does anyone know if i did the 2nd problem right?

The final answer is right, but your notation could be confusing, better to
write it as:

dy/dx=(3 x^2)/(x^3+1).

The argument which leads to your solution however is confused. It should
go more like:

y=f(g(x)), where g(x)=x^3+1, and f'(x)=1/x.

Then by the chain rule:

dy/dx = f'(g(x)) g'(x) = [1/(x^3+1)] 3x^2 =(3x^2)/(x^3+1).

RonL