Find y' given..

Question

Find y' given..

luigi0210 · Answer

$$\LARGE y=ln(5x^2+y^2)$$

kainui · Answer

What's your best guess?

luigi0210 · Answer

Wait.. so do it like.. 
$$\LARGE =\frac{1}{5x^2+y^2}*10x+2yy'$$?

kainui · Answer

I wouldn't do it this way. I'd exponentiate both sides so you have:

$$e^y=5x^2+y^2$$

That way you don't have to deal with fractions so much in doing algebra solving for y'.

anonymous · Answer

$$y' =\frac{1}{5x^2+y^2}*10x+2yy'$$n is more like it

kainui · Answer

@satellite73 what?

kainui · Answer

Following my suggestion...$$e^y*y'=10x+2yy'$$$$y'(e^y-2y)=10x$$$$y'=\frac{ 10x }{ e^y-2y }$$

anonymous · Answer

$$y' = \LARGE =\frac{1}{5x^2+y^2}*(10x+2yy')$$

kainui · Answer

See none of you other people have found y' though lol.

anonymous · Answer

it's just a matter of isolating y'

loser66 · Answer

wonder why everybody help this guy but not help me??

kainui · Answer

Is your question a simple one about derivatives? =P

anonymous · Answer

@Loser66 you're at graduate level, we're not lol

loser66 · Answer

who said that? I am undergraduate student.

kainui · Answer

@Loser66 What's your question, I don't even see it anywhere.

anonymous · Answer

probably Advanced Linear Algebra

luigi0210 · Answer

@Kainui I understand what you were doing but it didn't accept the answer >.<

anonymous · Answer

dy/dx = (10x)/(5x^2+y^2 - 2y) did you have this?

kainui · Answer

I'm sorry computers are stupid, good luck figuring out how to mold the right answer into the answer it wants.

luigi0210 · Answer

Yea, I see where I messed up now

Thanks @sourwing and @Kainui :)

luigi0210 · Answer

And agreed @Kainui :)