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- anonymous

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- madhu.mukherjee.946

first find dy/dx

- madhu.mukherjee.946

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

- anonymous

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

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## More answers

- anonymous

Oooo my bad and then would i derive that again?

- madhu.mukherjee.946

yep

- anonymous

Im having trouble deriving it

- anonymous

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

- madhu.mukherjee.946

wait i made a mistake

- madhu.mukherjee.946

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

- madhu.mukherjee.946

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

- madhu.mukherjee.946

now do second time derivative

- 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

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

- anonymous

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

- anonymous

@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

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

- zepdrix

derivative of y' is y''

- 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

Oooo okay and the value of dy/dx is?

- anonymous

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

- anonymous

@zepdrix

- zepdrix

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

- anonymous

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

- 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

\[\large\rm 10+\color{royalblue}{(4y^3)'}y'+4y^3\color{royalblue}{(y')'}=0\]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

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

- anonymous

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

- 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

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

- zepdrix

woah woah woah :O

- zepdrix

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

- anonymous

Oops

- zepdrix

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

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