SSTjumbo From , joined Dec 1969, posts, RR: Posted (12 years 8 months 1 week 12 hours ago) and read 1459 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?

PanAm747 From United States of America, joined Feb 2004, 4242 posts, RR: 8
Reply 1, posted (12 years 8 months 1 week 11 hours ago) and read 1430 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!

Docpepz From Singapore, joined May 2001, 1969 posts, RR: 3
Reply 5, posted (12 years 8 months 6 days 11 hours ago) and read 1394 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.

JetService From United States of America, joined Feb 2000, 4798 posts, RR: 11
Reply 6, posted (12 years 8 months 6 days 3 hours ago) and read 1380 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.

Delta-flyer From United States of America, joined Jul 2001, 2676 posts, RR: 7
Reply 7, posted (12 years 8 months 5 days 20 hours ago) and read 1369 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.

KROC From , joined Dec 1969, posts, RR:
Reply 8, posted (12 years 8 months 5 days 20 hours ago) and read 1368 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...

Bernard Shakey From United States of America, joined Oct 2001, 560 posts, RR: 9
Reply 9, posted (12 years 8 months 5 days 20 hours ago) and read 1365 times:

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

Mindless drifter on the road, Carries such an easy load

JetService From United States of America, joined Feb 2000, 4798 posts, RR: 11
Reply 12, posted (12 years 8 months 5 days 11 hours ago) and read 1342 times:

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