i forgot what this differential was call

mattheman said:
i can do the 1st one just cant get my head around the last 2.
Click to expand...

ah, for the second one, expand the brackets, and use the same method as the first. for the third one, dy/dx1 is effectively the same as d/dx of 4x+3/c where c is some number, so the result is 4/c. c in this case is x2-2. for dy/dx2, it is effectively d/dx of c/x-2 so you are going to end up with -c/(x2-2)^2, in this case c is 4x1+3.
therefore,
dy/dx1 is 4/x2-2
dy/dx2 is -4x1+3/(x2-2)^2
EDIT: Beaten ;)
 
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Vixen said:
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.
Click to expand...

Agreed, except slight picky point, the brackets won't expand to a quadratic, as the highest-order term will still be linear in both cases (eg x1x2 is first-order, x1(x2)^2 would be quadratic in x2.)

EDIT: glad lord_filbuster agrees with my maths.

EDIT EDIT: I'll stop excesssive editing and go sleep!
 
RoysonBobson said:
Agreed, except slight picky point, the brackets won't expand to a quadratic, as the highest-order term will still be linear in both cases (eg x1x2 is first-order, x1(x2)^2 would be quadratic in x2.)

EDIT: glad lord_filbuster agrees with my maths.

EDIT EDIT: I'll stop excesssive editing and go sleep!
Click to expand...

Yes, sorry I should have said "as if it were a quadratic".
 
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