Question

dy/dx= (3y-2x)/(8y-3x). Find d^2y/dx^2.

Answer 1

I understand that I am finding the second derivative, and I was wondering if I am able to derive it as 3y/8y - 2x/3x.

Answer 2

Noooo :O
Can't split up denominators like that!

You could split it up like:\[\Large \frac{3y}{8y-3x}-\frac{2x}{8y-3x}\]But I don't think that's really going to help us :)

Answer 3

A fraction is sort of like a house :D
The numerator is all the furniture and stuff.
We can move the furniture around, but we can't separate the foundation.

Ok maybe that was a silly explanation lol

Answer 4

Looks like we need to apply the `Quotient Rule` for this one, yes?\[\Large y''\quad=\quad \frac{\color{royalblue}{(3y-2x)'}(8y-3x)-(3y-2x)\color{royalblue}{(8y-3x)'}}{(8y-3x)^2}\]

Answer 5

Would the derivative of (3y-2x) be (3dy/dx-2)? And similar fashion for (8y-3x)' ?

Answer 6

yes, very good.

Answer 7

They're probably going to want the final answer in terms of x and y, NOT in terms of x, y and y'.

So you'll have a bit of work to do after you find the derivative :(
Plugging in the y's from your original problem.

Answer 8

Ok, just a quick quesiton. When I plugged this into an online calculator *shame shame* it gave me that (3y-2x)'=-2. Is that true?

Answer 9

If you typed it into wolfram, lemme check really quick.
*checking*...

No it is not true.
Do you see the weird curly d's on the derivative operator?
They took a `partial` derivative instead of a normal derivative with respect to x.

It's something different that you'll learn about later. Doesn't really apply here though :)

Answer 10

Ok, so after multiplying everything throughout the numerator, I now need to plug in my original dy/dx into the d^2y/dx^2?

Answer 11

\[\Large y''\quad=\quad \frac{\color{orangered}{(3y'-2)}(8y-3x)-(3y-2x)\color{orangered}{(8y'-3)}}{(8y-3x)^2}\]Yah you'll need to plug your original setup into each of those `y'`s :o

Answer 12

You could plug before multiplying everything out, or after, it's up to you :D

Answer 13

Ok. So I have the point (3,2) that I need to use to find the value of y''. Can I plug in those values as soon as I substitute the y's?

Answer 14

Yah that would be a good idea.
Plug those values in as soon as possible.

You should be able to do some adding and subtracting making all the expansions much easier.

Answer 15

Well.. when I substitute in the values, I get 0/49 am I wrong, or does this mean there is a local max/min?

Answer 16

No it doesn't mean something like that :)
But hmm, I'm coming up with y''(3,2) = -14/49

Answer 17

I found it easier to calculate the `y'` on the side,
it works out to y'(3,2) = 0, yes?

Answer 18

\[\large y''(3,2)\quad=\quad \frac{(3\cdot0-2)(8\cdot2-3\cdot3)-(3\cdot2-2\cdot3)(8\cdot0-3)}{(8\cdot2-3\cdot3)^2}\]

Answer 19

Plugging in 0's for y's, and the rest of the values, I get something like that ^

Answer 20

That first term in the numerator should give you something besides zero :o

Answer 21

Doh! I left out the -2 in 3(y')-2

Which would have given me -14/49

Answer 22

Ok. The question asks if the curve has a local max, min, or neither at the point. How do I determine that?

Answer 23

Ummmmmmmmmmmmmmm trying to remember...

Answer 24

There's a formula for second derivative test, i'm trying to remember what it looks like.

Answer 25

While you think, I'm just going to throw this out there. When I used the calculated y'(8y-3x) = -3 (and similarly for y'(3y-2x) via wolfram alpha and then evaluated, I came up with the same answer. Is that just coincidence?

Answer 26

I'm not sure what you calculated there.
Can you paste the link from wolfram?

Answer 27

http://www.wolframalpha.com/input/?i=derivative+of+8y-3x

Answer 28

Also, it's similarly used at http://www.derivative-calculator.net/#expr=%283y-2x%29%2F%288y-3x%29&showsteps=1

Answer 29

I'm referring to earlier when I asked, "Ok, just a quick quesiton. When I plugged this into an online calculator *shame shame* it gave me that (3y-2x)'=-2. Is that true?"

Answer 30

Sorry if I'm getting confusing here. I'm a little tired ;)

Answer 31

Mmm ok I'm trying to see what's going on here..

Answer 32

Ya I can't figure out why all these sites are calculating the partial derivative :(
hmm have you learned about partial derivatives at this point? :o

It's something you learn about in multi-variable calculus

Answer 33

Not that I know of! That's why I was asking.. I was afraid I may have dozed off when a shortcut was thought!

Answer 34

taught* rather

Answer 35

bah i dunno :3 now you're making me tired lol

Answer 36

Well, it's 1 a.m. you have the right to be tired! But any last thoughts on the local min, max, or neither bit? If not I'll just leave it for the first couple of minutes when I get to class :P

Answer 37

does the point have a max, min or neither,

I guess since y'(3,2) = 0, we can determine that it IS a critical point and may be a max or min.

Answer 38

And Ummmm let's see..

Answer 39

There is a second derivative test,
which would tell us if the function is concave up or down at the given point.

~If it's concave up, then the point is a minimum)
~If it's concave down, the point is a maximum)
~I guess it could also be neither, just a critical point that an inflection point. hmm,

Answer 40

Concavity Theorem: If the function f is twice differentiable at x=c, then the graph of f is concave upward at (cf(c)) if f(c)0 and concave downward if f(c)0.

Answer 41

Woops your inequality signs didn't show up when pasting :( lol

Answer 42

Does this mean that since y"=-17/49 it's concavity is down?

Answer 43

Yes that's what it should mean! :)

Answer 44

So it's a local max

Answer 45

Ya that sounds right.. i think! :) lol

Answer 46

Sounds good to me! Thanks a ton for the help!

Answer 47

np \c:/