1. ## implicit darivations.

yep, yall are handy... love ya's much.

So, it asks for us to find the y'' of x2 + xy + y3 = 1 implicitly.

Pointers?

2. I'm assuming you meant "Find $\displaystyle y''(x)$ where

$\displaystyle x^{3}+xy+y^{2}=1.$"

Is that correct? If so, what would be the first step?

3. yes thats correct, so i just take it like i would a regular derv?

4. Yes. Rather than do that problem for you, let me find the second derivative of y when $\displaystyle y^2- e^x+ xy- 2y= 4$.

Differentiating once, $\displaystyle 2yy'- e^x+ y+ xy'- 2y'= 0$. Differentiating again, $\displaystyle 2y'y'+ 2yy''- e^x+ y'+ y'+ xy''- 2y''= 2(y')^2+ 2yy''- e^x+ 2y'+ xy''- 2y''= 0$. Now, you can solve for y'':
$\displaystyle y''= \frac{e^x- 2(y')^2- 2y'}{2y+ x- 2}$.

You could, now, replace y' from the first derivative but typically, that is not necessary.

5. if im lookin for y'' i do nothing to the x's?

6. What do you mean by "do nothing to the x's"?

7. I must be missing something...
You never took the derivative of the x's.

8. HallsofIvy did do the differentiation correctly, taking into account that y = y(x). Let's say I have the term xy. Its derivative w.r.t. x is y + xy'. Do you see that?

9. 6x+2=y''(x)... is that right?

10. Originally Posted by FreaKariDunk
6x+2=y''(x)... is that right?
From where did that come?

11. With a reaction like that, it must be wrong. I don't know what I'm missing. I'm gonna re-read the section and re-explore that problem.

12. Take a good look at HallsofIvy's Post # 4. Start with the same equation, and see if you can't reproduce those results.

13. as i retry this... Im workin on finding the formula for dy/dx
and i got to here... 3x^2+[x(dy/dx) + y(dx/dx)]+2y=0

now, the stuff in the brackets.. how do I simplify that... in the first term can I 'cancel' the x leaving y in the numerator. but then that leaves the second term...

14. No, you can't cancel inside the brackets. The symbol $\displaystyle dy/dx$ is one symbol. It is not a fraction. However, the $\displaystyle dx/dx = 1$, so that simplifies a bit. Incidentally, I would agree with your differentiation except for the last term. You should use the chain rule on it thus:

$\displaystyle 3x^{2}+[xy'+y]+2yy'=0.$

Can you continue?

15. So now I have to solve for y'...
so it would be (-3x^2-y)/(x2y) ?

Page 1 of 2 12 Last