SSTjumbo
Topic Author
Posts: 2579
Joined: Sun Jun 17, 2001 3:29 am

### Hey Everybody, It's Time To Help Me With Calc!

If y=f(x) and m=Dx/Dy, how do I find Dm/Dy? For instance, say y=x^2+3x+2 making Dy/Dx=2x+3 => m=2x+3, how would I find Dm/Dy? Does it involve natural logs maybe?
I don't know, so this is my signature.

PanAm747
Posts: 4713
Joined: Mon Mar 01, 2004 4:46 am

### RE: Hey Everybody, It's Time To Help Me With Calc!

Dy=2x+3

Dm=2

Dm/Dy=(2)/(2x+3)

I believe you have two equations and you just divide them:

1. y=f(x)=x^2+3x+2=
Dy=2x+3

2. m=f(x)=2x+3
Dm=2

Then:

Dm/Dy=(2)/(2x+3)
Pan Am:The World's Most Experienced Airline - P(oor) S(ailor's) A(irline): San Diego's Hometown Airline-Catch Our Smile!

Guest

### RE: Hey Everybody, It's Time To Help Me With Calc!

No, this is what it is...

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-/EP'IFHwp[evgjwrpsvg

pdihnifjvw[PG
23-4098T-3
'PGJP

Just holla holla if you need some more help.

docpepz
Posts: 1706
Joined: Thu May 24, 2001 8:20 pm

### RE: Hey Everybody, It's Time To Help Me With Calc!

You do have to realise that dy/dx is NOT a division, even though in many cases it may seemingly look like one.

anyway you said that m = dx/dy. y=x^2+3x+2 making Dy/Dx=2x+3 => m=2x+3

why is m = 2x + 3? because if dy/dx = 2x +3, m is not 2x + 3, since you defined m as dx/dy. which makes m
1/(2x+3)

unless it's a typo on your part. However, assuming it's not a typo:

dm/dy = (dx/dy) x (dm/dx)

dy/dx = 2x + 3
dx/dy = m = 1/(2x+3)
dm/dx = 2/(2x+3)xY2

multiply dx/dy by dm/dx to get the answer.

By this i am assuming that you correctly defined m as dx/dy and not dy/dx.

SSTjumbo
Topic Author
Posts: 2579
Joined: Sun Jun 17, 2001 3:29 am

### RE: Hey Everybody, It's Time To Help Me With Calc!

OOPS, I meant m=Dy/Dx!!! my bad
I don't know, so this is my signature.

docpepz
Posts: 1706
Joined: Thu May 24, 2001 8:20 pm

### RE: Hey Everybody, It's Time To Help Me With Calc!

SSTjumbo: even if you meant dx/dy, the question can still be done, and is interesting. You should solve the question the same way.

(dm/dy) = (dx/dy) x (dm/dx)

You should still solve it that way. Except, since m is defined as dy/dx, you should define m, and use the formula (dx/dy) = (inverse of dy/dx) to solve it.

Hope it helps.

and regarding what PanAm747 said, I may be wrong,

but dy= 2x+3 dx, and not 2x+3.

It is wrong to write dy=2X+3 alone.

thus dy/dx= 2X+3.

It's quite important we learn not to forget the 'dx' term, especially in integration. Have you learnt integration by parts or by substitution? If you have, you'll know why.

JetService
Posts: 4611
Joined: Mon Feb 21, 2000 1:12 pm

### RE: Hey Everybody, It's Time To Help Me With Calc!

Simplify this way to find Dm/Dy..
(to find Dm/Dy, assuming y=f(x) and m=Dx/Dy...
using variable 'S')
Muliply both sides by 'd'
S=d(y)/fm(x/y)

Now variable use 'S' again to bring 1/m over

SS=Dm(y)/f(x/y)

Simplify 'y' (using variable 'k')

SS=Dm/fk(x)

Now using your 'for instance' result (=> m=2x+3), reduce 'x' by variable 'T' divided by (b/c) and you get the answer.

SST = Dmb/Fck

Does it involve natural logs maybe?

Save your natural logs for Sunday Morning.

delta-flyer
Posts: 2631
Joined: Mon Jul 30, 2001 9:47 am

### RE: Hey Everybody, It's Time To Help Me With Calc!

SSTjumbo,

I think you have misstated the problem. I suspect the question is to solve for dm/dx, which is nothing but the second derivative of y with respect to x.

If the problem is correctly stated, then you must first find x as a function of y.

Since m = dy/dx, you can easily get that by differentiating f(x). Next, substitute for x so you get m as a function of y, not x. Then you can differentiate with respect to y to get dm/dy. But the problem is that getting x as a function of y gets tough if y=f(x) is greater than a first order equation, as in your example.

Cheers,
Pete

"In God we trust, everyone else bring data"

KROC
Posts: 18919
Joined: Mon May 08, 2000 11:19 am

### RE: Hey Everybody, It's Time To Help Me With Calc!

What do you mean "Hey Everybody, It's Time To Help Me With Calc!" You make it sould like we have no chice. We help you or die. Well son, here is my offering...

http://www.casio.com/calculators/product.cfm?section=24&market=0&product=1865

Bernard Shakey
Posts: 542
Joined: Thu Oct 18, 2001 4:05 am

### RE: Hey Everybody, It's Time To Help Me With Calc!

With all due respect to SST, Jetservice that was classic.

SSTjumbo
Topic Author
Posts: 2579
Joined: Sun Jun 17, 2001 3:29 am

### RE: Kroc

That's right, you have to help me or else I'll kill you.
I don't know, so this is my signature.

SSTjumbo
Topic Author
Posts: 2579
Joined: Sun Jun 17, 2001 3:29 am

### RE: Jetservice

SST = Dmb/Fck

I don't know, so this is my signature.

JetService
Posts: 4611
Joined: Mon Feb 21, 2000 1:12 pm

### RE: Hey Everybody, It's Time To Help Me With Calc!

LOL, SST!!!!! I can't! That batteries in my abacus just ran dry.

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