anonymous one year ago ...

first find dy/dx

dy/dx=10x+4y^3d/dx

3. anonymous

Okay so it would be -5x/2y^3

4. anonymous

Oooo my bad and then would i derive that again?

yep

6. anonymous

Im having trouble deriving it

7. anonymous

Would it be10+12y^2d/dx?

it would be dy/dx=10x+4y^3dy/dx

dy/dx-4y^3dy/dx=10x dy/dx(1-4y^3)=10x

now do second time derivative

12. zepdrix

Would it be10+12y^2dy/dx? Do you mean for your second derivative? Hmm, no. It's going to be a little more complicated than that :) You need to product rule the y^3 and y'

13. zepdrix

$\large\rm 10x+4y^3y'=0$$\large\rm 10+\color{royalblue}{(4y^3)'}y'+4y^3\color{royalblue}{(y')'}=0$Product rule! :)

14. anonymous

Okay so it would be 10 + 12y^2+4y^3?

15. anonymous

@zepdrix

16. zepdrix

$\large\rm 10+\color{royalblue}{(4y^3)'}y'+4y^3\color{royalblue}{(y')'}=0$$\large\rm 10+\color{orangered}{(12y^2)}y'+4y^3\color{orangered}{(y'')}=0$Hmm I don't see your derivative y' anywhere D:

17. anonymous

Would the derivative of y' just be d? That part confuses me

18. zepdrix

derivative of y' is y''

19. anonymous

5x^2+y^4=-9 diff. with respect to x $10x+4y^3 \frac{ dy }{ dx }=0 ...(1)$ $4y^3 \frac{ dy }{ dx }=-10x,\frac{ dy }{ dx }=\frac{ -5x }{ 2y^3 }$diff. (1)w.r.t x again $10+4y^3\frac{ d^2y }{ dx^2 }+12y^2\left( \frac{ dy }{ dx } \right)^2=0$ put the value of dy/dx and find $\frac{ d^2y }{ dx^2 }$ and finally plug the values of x and y

20. anonymous

Oooo okay and the value of dy/dx is?

21. anonymous

I got 10+4d^2y/dx^2+12=0 so 26 + dy/dx=0? So it would be 26?

22. anonymous

@zepdrix

23. zepdrix

I dunno, there is too much stuff going on here :( Lemme get organized...

24. anonymous

Okay sorry for bothering but this is the last question I need and I really don't get it

25. zepdrix

If you like using the d/dx's you can do that :) But I'm gonna be lazy and use the prime notation, it's just so much "neater" to work with.$\large\rm 5x^2+y^4=-9$Taking first derivative, applying chain rule to the y,$\large\rm 10x+4y^3y'=0$Evaluate this at x=2, y=1. $\large\rm 10(2)+4(1)^3y'=0\qquad\to\qquad \color{green}{y'=-5}$Stick that green equation into your back pocket, save it for later. Taking second derivative, we have to apply product rule,

26. zepdrix

$\large\rm 10+\color{royalblue}{(4y^3)'}y'+4y^3\color{royalblue}{(y')'}=0$This is just me "setting up" the product rule, that's what the blue is there for. It changes to orange when I've taken the derivative,$\large\rm 10+\color{orangered}{(12y^2y')}y'+4y^3\color{orangered}{y''}=0$Ok with those two derivatives? :o They're a little tricky.

27. zepdrix

Aw mannnn, your brain esplode? :( Oh noessss, brain everywhere.. ahhh what a mess!

28. anonymous

Ok haha its a little confusing but I'm following you

29. zepdrix

So I setup the product rule between the $$\rm 4y^3$$ and the $$\rm y'$$ The $$\rm 4y^3$$ gave us $$\rm 12y^2y'$$ while the $$\rm y'$$ gave us $$\rm y''$$.$\large\rm 10+(12y^2\color{green}{y'})\color{green}{y'}+4y^3y''=0$We know that y'=-5 when x=2, y=1. Remember? We stuck it in our back pocket. Let's pull it out, along with the other information they gave us, and plug it all in. $\large\rm 10+(12(1)^2\color{green}{(-5)})\color{green}{(-5)}+4(1)^3y''=0$And we don't plug anything in for y'', because that's what we're trying to solve for! :)

30. anonymous

So 10+(12*-5)(-5+4)=0 10-60(-1)=0 50?

31. zepdrix

woah woah woah :O

32. zepdrix

That -5 shouldn't be with the 4, not sure how that happened :O

33. anonymous

Oops

34. zepdrix

Do you have to enter it into an online thing or no? :o