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anonymous
 one year ago
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anonymous
 one year ago
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madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.2first find dy/dx

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.2dy/dx=10x+4y^3d/dx

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay so it would be 5x/2y^3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oooo my bad and then would i derive that again?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im having trouble deriving it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would it be10+12y^2d/dx?

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.2wait i made a mistake

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.2it would be dy/dx=10x+4y^3dy/dx

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.2dy/dx4y^3dy/dx=10x dy/dx(14y^3)=10x

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.2now do second time derivative

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2`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
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm 10x+4y^3y'=0\]\[\large\rm 10+\color{royalblue}{(4y^3)'}y'+4y^3\color{royalblue}{(y')'}=0\]Product rule! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay so it would be 10 + 12y^2+4y^3?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm 10+\color{royalblue}{(4y^3)'}y'+4y^3\color{royalblue}{(y')'}=0\]\[\large\rm 10+\color{orangered}{(12y^2)}y'+4y^3\color{orangered}{(y'')}=0\]Hmm I don't see your derivative y' anywhere D:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would the derivative of y' just be d? That part confuses me

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2derivative of y' is y''

anonymous
 one year ago
Best ResponseYou've already chosen the best response.05x^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
 one year ago
Best ResponseYou've already chosen the best response.0Oooo okay and the value of dy/dx is?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got 10+4d^2y/dx^2+12=0 so 26 + dy/dx=0? So it would be 26?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2I dunno, there is too much stuff going on here :( Lemme get organized...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay sorry for bothering but this is the last question I need and I really don't get it

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2If 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
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm 10+\color{royalblue}{(4y^3)'}y'+4y^3\color{royalblue}{(y')'}=0\]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
 one year ago
Best ResponseYou've already chosen the best response.2Aw mannnn, your brain esplode? :( Oh noessss, brain everywhere.. ahhh what a mess!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok haha its a little confusing but I'm following you

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So 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
 one year ago
Best ResponseYou've already chosen the best response.0So 10+(12*5)(5+4)=0 1060(1)=0 50?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2That 5 shouldn't be with the 4, not sure how that happened :O

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Do you have to enter it into an online thing or no? :o
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