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Hey Everybody, It's Time To Help Me With Calc!  
User currently offlineSSTjumbo From , joined Dec 1969, posts, RR:
Posted (12 years 9 months 3 weeks 5 days 15 hours ago) and read 1488 times:

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?

12 replies: All unread, jump to last
 
User currently offlinePanAm747 From United States of America, joined Feb 2004, 4242 posts, RR: 8
Reply 1, posted (12 years 9 months 3 weeks 5 days 14 hours ago) and read 1459 times:

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!
User currently offlineBoeing757fan From , joined Dec 1969, posts, RR:
Reply 2, posted (12 years 9 months 3 weeks 5 days 13 hours ago) and read 1455 times:

No, this is what it is...

weopfjW+EPIJHSWIFJP
-/EP'IFHwp[evgjwrpsvg

pdihnifjvw[PG
23-4098T-3
'PGJP

 Big thumbs up

Just holla holla if you need some more help.



User currently offlineDocpepz From Singapore, joined May 2001, 1971 posts, RR: 3
Reply 3, posted (12 years 9 months 3 weeks 5 days 13 hours ago) and read 1450 times:

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.


User currently offlineSSTjumbo From , joined Dec 1969, posts, RR:
Reply 4, posted (12 years 9 months 3 weeks 4 days 14 hours ago) and read 1432 times:

OOPS, I meant m=Dy/Dx!!! my bad

User currently offlineDocpepz From Singapore, joined May 2001, 1971 posts, RR: 3
Reply 5, posted (12 years 9 months 3 weeks 4 days 14 hours ago) and read 1423 times:

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.


User currently offlineJetService From United States of America, joined Feb 2000, 4798 posts, RR: 11
Reply 6, posted (12 years 9 months 3 weeks 4 days 5 hours ago) and read 1409 times:

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  Big grin

Does it involve natural logs maybe?

Save your natural logs for Sunday Morning.



"Shaddap you!"
User currently offlineDelta-flyer From United States of America, joined Jul 2001, 2676 posts, RR: 6
Reply 7, posted (12 years 9 months 3 weeks 3 days 23 hours ago) and read 1398 times:

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



User currently offlineKROC From , joined Dec 1969, posts, RR:
Reply 8, posted (12 years 9 months 3 weeks 3 days 23 hours ago) and read 1397 times:

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


User currently offlineBernard Shakey From United States of America, joined Oct 2001, 560 posts, RR: 9
Reply 9, posted (12 years 9 months 3 weeks 3 days 22 hours ago) and read 1394 times:

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


Mindless drifter on the road, Carries such an easy load
User currently offlineSSTjumbo From , joined Dec 1969, posts, RR:
Reply 10, posted (12 years 9 months 3 weeks 3 days 14 hours ago) and read 1377 times:

That's right, you have to help me or else I'll kill you.  Acting devilish  Acting devilish  Acting devilish

User currently offlineSSTjumbo From , joined Dec 1969, posts, RR:
Reply 11, posted (12 years 9 months 3 weeks 3 days 14 hours ago) and read 1374 times:

SST = Dmb/Fck

Please mathematically prove this.  Big thumbs up


User currently offlineJetService From United States of America, joined Feb 2000, 4798 posts, RR: 11
Reply 12, posted (12 years 9 months 3 weeks 3 days 14 hours ago) and read 1371 times:

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




"Shaddap you!"
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