i forgot what this differential was call

I can't read what you're saying. use brackets and equals signs etc and you might have more luck getting an answer. I have a couple of ideas.

Or better still, draw it, scan it in and upload and link :D

EDIT: Sorry if i sound mean. I'd like to be able to answer the question.

Sin_Chase said:
my guess would be: Higher order differential
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mattheman said:
the 1 and 2 are at the bottom they are not power, i forgot what this differential was call
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Doubt it's a second-order differential. That's why I'm interested!
 
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i can understand the 1st one but have no idea how to do the 2nd or 3rd.

do i use the pruduct rule then eliminate x1 x2

do i use the quotent rule then eliminate x1 x2
 
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That's just saying x1 and x2 are different variables. Think of them as completely indepedant quantities. eg, when dy/dx1, treat x2 as constant. The rest is just straightforward algebraic manipulation and then differentiation.

As for the level, that would be AS level maths (Core 1) for the first two, and A2 (Core 4) for the bottom one.

EDIT: I was only joking about scanning it in, but thank you. It makes life so much easier! Hope my explanation helped.

EDIT EDIT: As below, you could think of these as partial differentiation, it's much the same thing. Yum yum. Next step: vector calculus :D
 
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RoysonBobson said:
That's just saying x1 and x2 are different variables. Think of them as completely indepedant quantities. eg, when dy/dx1, treat x2 as constant. The rest is just straightforward algebraic manipulation and then differentiation.

As for the level, that would be AS level maths (Core 1) for the first two, and A2 (Core 4) for the bottom one.

EDIT: I was only joking about scanning it in, but thank you. It makes life so much easier! Hope my explanation helped.

EDIT EDIT: As below, you could think of these as partial differentiation, it's much the same thing. Yum yum. Next step: vector calculus :D
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any chance of showing me :D
 
What Royson said, if it makes it easier rename them, so you have dy/dz and dy/dp, then just differentiate with respect to the z or the p, good luck!
 
lord filbuster said:
ok, Say you have a function F, and F=x^2 + 2y + xy, where x and y are just some variables, like x1 and x2.
then, df/dy is equal to 2+x, and df/dx is equal to 2x+y.
In both cases, the variable not being used in the differential is kept constant, and treated as just a number.
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i can do the 1st one just cant get my head around the last 2.
 
For the second one, expand the brackets out, it'll be quicker than using the product/chain rule.

For the bottom one you'll have to give me a minute. I avoid fractions :P That's what databooks are for!

EDIT:

for b) (2x1 +3)(x2 - 2) = 2x1x2 + 3x2 - 4x1 - 6 so dy/dx1 = 2(x2) - 4 & dy/dx2 = 2x1 + 3

for c) dy/dx1 = 4 / (x2 - 2) & dy/dx2 = -(4x1 + 3) / ( (x2 - 2)^2 ) - need to use chain rule on that bad boy

EDIT EDIT: it's past my bedtime. I'll check these tomorrow morning but I reckon they're right. Not entirely sure about the last one but I'm off to bed. Maybe someone else can check over the maths.
 
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RadioheadR6 said:
The question being is why post complex mathmatical formulae on this site - I would imagine that there are forums for this on the interweb?...
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There are plenty of people on this forum who can answer the questions he was given (which are not at all complex!). Of course it would be much easier if he actually asked them in English :p.

Partial derivatives are really easy, just differentiate with respect to whatever is on the bottom - ie dy/dx is x, dy/dz is z etc. The brackets need to be multiplied out as a quadratic and will then be very easy. As for the fraction, write it as numerator x 1/denominator.
 
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