anonymous
  • anonymous
...
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
madhu.mukherjee.946
  • madhu.mukherjee.946
first find dy/dx
madhu.mukherjee.946
  • madhu.mukherjee.946
dy/dx=10x+4y^3d/dx
anonymous
  • anonymous
Okay so it would be -5x/2y^3

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Oooo my bad and then would i derive that again?
madhu.mukherjee.946
  • madhu.mukherjee.946
yep
anonymous
  • anonymous
Im having trouble deriving it
anonymous
  • anonymous
Would it be10+12y^2d/dx?
madhu.mukherjee.946
  • madhu.mukherjee.946
wait i made a mistake
madhu.mukherjee.946
  • madhu.mukherjee.946
it would be dy/dx=10x+4y^3dy/dx
madhu.mukherjee.946
  • madhu.mukherjee.946
dy/dx-4y^3dy/dx=10x dy/dx(1-4y^3)=10x
madhu.mukherjee.946
  • madhu.mukherjee.946
now do second time derivative
zepdrix
  • 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'
zepdrix
  • zepdrix
\[\large\rm 10x+4y^3y'=0\]\[\large\rm 10+\color{royalblue}{(4y^3)'}y'+4y^3\color{royalblue}{(y')'}=0\]Product rule! :)
anonymous
  • anonymous
Okay so it would be 10 + 12y^2+4y^3?
anonymous
  • anonymous
@zepdrix
zepdrix
  • 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:
anonymous
  • anonymous
Would the derivative of y' just be d? That part confuses me
zepdrix
  • zepdrix
derivative of y' is y''
anonymous
  • 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
anonymous
  • anonymous
Oooo okay and the value of dy/dx is?
anonymous
  • anonymous
I got 10+4d^2y/dx^2+12=0 so 26 + dy/dx=0? So it would be 26?
anonymous
  • anonymous
@zepdrix
zepdrix
  • zepdrix
I dunno, there is too much stuff going on here :( Lemme get organized...
anonymous
  • anonymous
Okay sorry for bothering but this is the last question I need and I really don't get it
zepdrix
  • 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,
zepdrix
  • 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.
zepdrix
  • zepdrix
Aw mannnn, your brain esplode? :( Oh noessss, brain everywhere.. ahhh what a mess!
anonymous
  • anonymous
Ok haha its a little confusing but I'm following you
zepdrix
  • 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! :)
anonymous
  • anonymous
So 10+(12*-5)(-5+4)=0 10-60(-1)=0 50?
zepdrix
  • zepdrix
woah woah woah :O
zepdrix
  • zepdrix
That -5 shouldn't be with the 4, not sure how that happened :O
anonymous
  • anonymous
Oops
zepdrix
  • zepdrix
Do you have to enter it into an online thing or no? :o

Looking for something else?

Not the answer you are looking for? Search for more explanations.