SSTjumbo From , joined Dec 1969, posts, RR: Posted (13 years 10 months 2 weeks 5 hours ago) and read 2086 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 (13 years 10 months 2 weeks 4 hours ago) and read 2057 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)

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Docpepz From Singapore, joined May 2001, 1971 posts, RR: 3
Reply 5, posted (13 years 10 months 1 week 6 days 4 hours ago) and read 2021 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: 10
Reply 6, posted (13 years 10 months 1 week 5 days 20 hours ago) and read 2007 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: 6
Reply 7, posted (13 years 10 months 1 week 5 days 13 hours ago) and read 1996 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 (13 years 10 months 1 week 5 days 13 hours ago) and read 1995 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, 561 posts, RR: 8
Reply 9, posted (13 years 10 months 1 week 5 days 13 hours ago) and read 1992 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: 10
Reply 12, posted (13 years 10 months 1 week 5 days 4 hours ago) and read 1969 times:

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