Product rule :calculus?
87883
First write y as a product instead of a quotient:\n\n(2+x)*(2-3x)^(-1)\n\nNow you should be able to do this. \nUse 2+x as your f and (2-3x)^(-1) as your g. \nYou will have to use the chain rule to find g'.
549551
The way this is written, there is nothing to solve because it's not an equation.\n\nTo use the zero product property, you want to factor an expression that equals zero. Then you conclude that if a product equals zero, one of its factors must be zero. To factor the expression in your problem, combine any like terms to simplify, then factor the simplified expression.\n\nHope that helps.
435388
damnt i should have paid attention calculus...\n\nlet pam...pam...breat....dimension......carry the one.....as x approaches zero......carry  the two....hold on...\n\nthis could take a while.
398761
the product contains the expression (x-x) which = 0 thus the product is 0
525387
If you have two measurements X and Y with uncertanities (errors) Dx and Dy:\n\nthe uncertanity (error) for the product X*Y will be\nD1 = Y*Dx + X*Dy\n\nthe uncertanity (error) for the division X/Y will be\nD2 = X/Y * SQR((Dx/X)^2+(Dy/Y)^2)\n\nSearch for 'error propagation' on the internet.
1148067
You elevate the number to the power 2 like this:\n\n                                    x^2\n\nYou should still do the normal product but it is easier to thing you et the square of a certain number and most calculators already have the elevate to the second power function integrated\n\nCynthia
983225
remember the rule nx^n-1\n2x^2-1\n \nSo in this case it is 2x
788940
Dear friend,\n\nYour product is \n\n-5*(x^2) + 8*x -10*x + 16 =\n                           \n-5*(x^2) - 2*x + 16
441742
Assume there is a pair a_i, b_j with a non zero product.\nWrite fg= (a_i x^i + other terms) (b_j x^j + other terms) =\na_i b_j x^(i+j) is non zero.\n\nContradiction.
888343
Use the chain rule:  dy/dx = d/dx[(ln x)^x] = x(ln x)^(x-1)d/dx[ln x]
725168
Use L'Hospitals rule\nd(sin ax)/d( bx)\na*cos(ax)/b lim x----> 0  = a/b
192120
Use the chain rule. Let f(y)=y^(1/2) and g(x)=x-2. Then your expression is just f(g(x)), and the chain rule states that D(f(g(x))) = Df(g(x))*Dg(x) = [(1/2)*(g(x))^(-1/2)] * [g'(x)] = [(1/2)*(x-2)^(-1/2)] * [1].\nI suggest doing several more of these questions until you no longer need to explicitly write out f and g, but can do so in your head.
635930
x+2/x-2 \n= (x^2 - 2x + 2)/x [BODMAS rule]\nSo Eric is correct.
1041171
f'(x) = 11ln(5x^2+7x) + 11x(10x+7)/(5x^2+7x) using the product rule.\n\nf'(2) = 11ln(5*2^2+7*2) + 11(2)(10*2+7)/(5*2^2+7*2)\n= 11ln(34) + 594/34\n= 56.260554006071892933393257662056
742785
as x ->pi/4 (1-tan x) ->0 and sec x is not infinite(sqrt(2))\n\nso limit is product =0
482671
I don't think the concept you are talking about goes by the name of "Simpson's rule", not sure what its called but it should be (MO + ML*4 + MP)/6 or with your example (2+4*4+6)/6=4\n\nSimpson's rule on the other hand has to do with approximating integrals (the area under a curve).  Which unfortunately its been a few years and I haven't retained that information.\n\nAnd maybe I'm wrong and there are 2 Simpson's rules, it's possible...
805634
It is easier (by far) to do this with logarithmic differentiation:\ndy/dx=y*d/dx[ln y].\n\nHere, with y equal to your expression,\nln y=ln 10 +ln x +4ln(x-1)\n-3ln(x-2)-2ln(x+1),\nso the derivative of the logarithm is\n1/x +4/(x-1) -3/(x-2)-2/(x+1).\nThe first derivative is now your original expression times this.\n\nTo find the second derivative, use the product rule. The derivative of y is already known, and the second derivative of the logarithm is easy.