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first find dy/dx
dy/dx=10x+4y^3d/dx
Okay so it would be -5x/2y^3

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

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