Question

find d^2y/dx^2 if x^3y^2 = 5

i can find the dy/dx but i have no idea how to find d^2y/dx^2

Answer 1

Hmmm c:

Answer 2

\[\large 3x^2y^2+2x^3yy'=0\]\[\large y'=\frac{ -3x^2y^2 }{ 2x^3y }\]\[\large y'=\frac{ -3y }{ 2x }\]
Just to make sure you're on the right track, is this what you got for y'?
(y' is the same as dy/dx, just in case there's any confusion about that) :)

Answer 3

yep, i got that. :)

Answer 4

So from here, taking the derivative with respect to x AGAIN, will give us y'' on the left, anddddd looks like we'll have to apply the quotient rule on the right. :O Need help setting it up?

Answer 5

yes please!

Answer 6

\[\large (\frac{ u }{ v })'=\frac{ u'v=uv' }{ v^2 }\]\[\large (\frac{ -3y }{ 2x })'=\frac{ (-3y)'(2x)-(-3y)(2x)' }{ (2x)^2 }\]

Ok here is our initial setup, the primes are the terms we need to take a derivative of.

Answer 7

That part make sense? :D Now we just need to remember, whenever we differentiate a y-term, that a y' will pop out as a result of the chain rule (and because we differentiated WRT x, not y).

Answer 8

yeah. i get the chain rule, but i still don't get the correct answer... i dont know why...

Answer 9

Ok you're prolly just missing a simple step, they want the final answer in terms of X and Y ONLY. So see how we end up with a y' somewhere in there? We need to PLUG IN our answer that we got for y' earlier.

Answer 10

2x * (-3\frac{ dx }{dy}) - (-3) * 2

is that right for the top part?

Answer 11

oops sorry, i'm new

Answer 12

hehe ^^

Answer 13

I dunno, yours is a little hard to read c: lemme know if this is what you're coming up with.
\[\large \frac{ -3y'(2x)-(-3y)2 }{ 4x^2 }=\frac{ 6y-6xy' }{ 4x^2 }\]

Answer 14

\[2x * 3\frac{ dx }{ dy } - 2 * (-3y)\]

Answer 15

all over 2x^2

Answer 16

Hmm where'd the minus go on the first term? :O
And in your denominator, make sure you're submitting it as (2x)^2,
Because 2x^2 is not the same thing :D just in case you're doing this on webassign or something like that.

Answer 17

but is that right? because the answer is definitely not that.

i'm practicing for a test i have tomorrow.

Answer 18

\[(-3y)'(2x)=(-3\frac{ dy }{ dx })(2x)\]
The minus disappeared from your first term here, careful nowww :O

Answer 19

oops, yeah. i missed that.

Answer 20

Earlier we had found the FIRST derivative to be this:
\[\large \frac{ dy }{ dx }=\frac{ -3y }{ 2x }\]
So to clean up our SECOND derivative, we should plug this in for dy/dx.

Answer 21

so is what i have the answer then?

Answer 22

I think this is the answer you want to try and get.

Answer 23

requires quite a bit of simplification :O

Answer 24

\[- \frac{ 3 }{ 2 } \left( x \frac{ dy }{ dx } -y \right)\]
then all the stuff inside the brakets over x^2 that's the answer given

Answer 25

Oh they didn't change the dy/dx to x's and y's? Oh i see, they were taking it easy on you XD interesting.. i went too far with that, my bad.

Answer 26

sec ill take another look :3

Answer 27

okie. thanks by the way for the help

Answer 28

\[\huge \frac{ 6y-6x \frac{ dy }{ dx } }{ 4x^2 }=\frac{ -6(x \frac{ dy }{ dx }-y) }{ 4(x^2) }\]

Answer 29

\[\huge =\frac{ -3 }{ 2 }(\frac{ x \frac{ dy }{ dx } -y}{ x^2 })\]

Answer 30

?!?!

Answer 31

Understand the steps I took? :O it's a lil tricky, bunch of stuff going on. A lil bit of factoring and such.

Answer 32

i'm trying to follow it. i get some of it... i don't know why all the stuff gets moved around though...

Answer 33

Were you able to get it to this form yet?
\[\huge \frac{ 6y-6x \frac{ dy }{ dx } }{ (2x)^2 }\]

Answer 34

\[-6x \frac{ dy }{ dx } + 6y\]

that's what i got so far... so yeah i guess it's the same if i move that 6y over to the other side...

Answer 35

Yah subtraction looks prettier if you write it on the right ^^ hehe

Answer 36

and i get how you get 4x^2 on the bottom. but what i'm now confused with is why the answer is broken up the way it is...

Answer 37

There is no reason for it, the way they broke it up is quite unnecessary. If you take a test, your teacher is very unlikely to take off any points for not simplifying it to that exact point.

You should probably notice that the top and bottom can have a 2 divided out of them. But that's the only simplification necessary. The rest was just your book being silly (or did teacher give you this problem in class?).

Answer 38

this was a practice test from the teacher. and that was the solution. i also have another questoin. how did the -6y get to just - y?

Answer 39

Understand how factoring works? :)