what are symmetric difference quotients in calculus?
957360
The symmetric difference quotient is\n\n[f(x+h) - f(x-h)]/2h.\n\nIt is the average of the right and left quotients [f(x+h) - f(x)]/h and [f(x) - f(x-h)]/h, where h > 0. So it tends to give a better approximation of the derivative.
733399
Come on, this is an EASY one.\n\nHere's how you do identities in general.  Look to see what the differences are in the two sides.  Then, look for an identity that can fix those differences.\n\nFor example, the main difference between the left and right side is, there is sin^2 on the left, but not on the right.  So somehow, the sin^2 has to disappear.  Additionally, there are only cos^2's on the right, so whatever you do you your sin^2, it has to turn into some form of cos^2.\n\n\n\nAre you GETTING this?\n\n\n\n\nStart with the left side.  Replace sin^2 (theta) with 1 - cos^2 (theta).  Do the algebra that ensues to get the right side.
1144759
I don't know the answer to your question, but here is a good website that might help:\n\nhttp://www.purplemath.com/modules/asymtote3.htm
110508
(x^2+5) / (x^3+15-2x^2+1)^2\non [0,1]
909284
((1/x) + (1/y)) / ((1/x) - (1/y))\n\n((y + x)/xy) / ((y - x)/xy)\n\n((y + x)/xy) * ((xy)/(y - x))\n\n(xy(y + x))/(xy(y - x))\n\n(y + x)/(y - x) = (x + y)/(-x + y) = (x + y)/(-(x - y)) = -(x + y)/(x - y)
1149867
We have x=rcos(θ), and y=rsin(θ).  \n\n\nTherefore x=4sin(θ)cos(θ) and y=4sin(θ)sin(θ).  \nAnd x^2+y^2= 16sin(θ)^2(cos(θ)^2+sin(θ)^2) =16sin(θ)=4y.  \nThus x^2+y^2-4y+4-4 =x^2+(y-2)^2-4=0 and x^2+(y-2)^2=4.\n\nAre you sure the second one is given by:\nρsin(θ)=2cos(θ) and not ρsin(φ)=cos(θ)?\n\nif it is ρsin(φ)=2cos(θ), then first notice that φ≠kπ for any integer k.\nThen we have ρ=2cos(θ)/sin(φ)\nNext we have x=ρcos(θ)sin(φ)=2cos(θ)^2, and y=ρsin(θ)sin(φ)=2cos(θ)sin(θ).\n\nThus x^2+y^2= 4cos(θ)^2(cos(θ)^2+sin(θ)^2) =2x,\nand x^2-2x+1-1+y^2= (x-1)^2-1+y^2=0,\nso (x-1)^2+y^2=1
252475
abc + d = 2\n\nMultiply by d, and re-arrange, you get:\n\n   d^2 - 2d = -abcd\n\nAdd one to both sides:\n\n   (d-1)^2 = 1 - abcd\n\nBy symmetry, this means that a-1, b-1, c-1, and d-1 must all be square roots of the same number.  If we let:\n\n   1 - abcd = u^2\n\nthen a = 1 +/- u,\n     b = 1 +/- u,\n     c = 1 +/- u,\n     d = 1 +/- u\n\nWe can look at essentially three different versions of this:\n\n    a = 1+u, b=1+u, c=1+u, d=1+u\n    solution: u=0, a=b=c=1 [all have u=0 solution]\n\n    a=1+u, b=1+u, c=1+u, d=1-u\n    solution: u=-2, a=b=c=-1, d=3\n\n    a=1+u, b=1+u, c=1-u, d=1-u\n    no non-zero solution
939642
Counterexample.  Let u(x), v(x) be members of {sin x, cos x}.  Both of those are solutions, but you do not get zero by adding one to the other, or by doubling one of them.
332712
The statement is true for :\nIf A is a skew-symmetric matrix, then A = - At (A transpose)\nSo (At)^(-1)= (A^(-1))t\nSo (-A)^(-1)= (A^(-1))t\nSo -1((A)^(-1)= (A^(-1))t\nSo A^(-1) is skew-symmetric
372419
I fear you have a missing element here... do you want the differencial of 4P^2..etc because if so you need to know what 4P^2..etc equals, or is 4P^2..etc the differencial of another expression, because if so your question is a bit like saying what is the answer to x=2\n\nIf you want to differencial of 4P^2..etc then I suggest you solve the original equation for t which will make the algebra easier and then use the chain rule where dP/dt = [1/(dt/dP)]
883038
If you mean Hermite see\nhttp://mathworld.wolfram.com/HermiteDifferentialEquation.html
532934
INT[(1 - x)/x dx]\nINT[x^-1 (1-x) dx]\nINT[x^-1 - 1 dx]\nln|x| - x + C
751318
x^2-x-6>0\nor, x^2-3x+2x-6>0\nor, x(x-3)+2(x-3)>0\nor, (x-3)(x+2)>0\nor x>3 and x<2\n\nhope this helps :)
787521
Example:\nIf z=f(u,v),\nu=g(x,y) and\nv=h(x,y), then\n(all derivatives are supposed to be partial derivatives)\ndz/dx=dz/du*du/dx+dz/dv*dv/dx\ndz/dy=dz/du*du/dy+dz/dv*dv/dy\n\nAlternatively, you can set of the matrices\n[[dz/du dz/dv]]\nand\n[[ du/dx du/dy]\n[dv/dx dv/dy]]\nand get that\n[[dz/dx dz/dy]]\nis the product of the two matrices.
1126229
Most of you I think misinterpretd the question(or maybe it was me), but I understood this:\n"is that function odd or even (referring to symmetry)?"\nWell if that is the question, do this:\nF(x) is even (symmetyric with respect to the y-axis) if F(-x)=F(x)\nF(x) is odd (symmetric with respect to the origin) if F(-x)= -F(x)\nThe function does have symmetry if it doesn't fill the above situations.\n\nIn this case by simple substitution the function is not symmetric at all.\n\nNOTE: There is a third type of symmetry for relations (not functions!) which is with respect to the x-axis.
1058665
|sin(z)|^2 + |cos(z)|^2\n\nWhere |f(z)|^2 is understood to mean f(z)*~f(z), where ~f(z) is the complex congugate of f(z): ~f(x + iy) = f(x - iy)\n\nWriting the real and imaginary parts of the arguments explicitly, we have:\n\n= |sin(x+iy)|^2 + |cos(x+iy)|^2\n\nUsing the sum of angles formulas:\n\n= |(sin(x)cos(iy) + cos(x)sin(iy)|^2 + \n  |(cos(x)cos(iy) - sin(x)sin(iy)|^2\n\nExpanding in terms of complex congugates:\n\n= ((sin(x)cos(iy) + cos(x)sin(iy))*(sin(x)cos(-iy) + cos(x)sin(-iy)) +\n  ((cos(x)cos(iy) - sin(x)sin(iy))*(cos(x)cos(-iy) - sin(x)sin(-iy))\n\nNoting that cos(-x) = cos(x) and sin(-x) = -sin(x):\n\n= ((sin(x)cos(iy) + cos(x)sin(iy))*(sin(x)cos(iy) - cos(x)sin(iy)) +\n  ((cos(x)cos(iy) - sin(x)sin(iy))*(cos(x)cos(iy) + sin(x)sin(iy))\n\n\nMultiplying this out (both terms are the product of a sum and difference of two terms, which yield the difference of squares):\n\n= (sin(x)cos(iy))^2 - (cos(x)sin(iy))^2 + (cos(x)cos(iy))^2 - (sin(x)sin(iy))^2\n\nFactoring:\n\n= ((sin(x))^2 + (cos(x))^2)*cos(iy) - ((sin(x))^2 + (cos(x))^2)*sin(iy)\n\nbut ((sin(x))^2 + (cos(x))^2) = 1, so\n\n= (cos(iy))^2 - (sin(iy))^2\n\nFrom Euler's formula (exp(ix) = cos(x) + i*sin(x)), we can show that:\n\ncos(iy) = (exp(-y) + exp(y))/2 = cosh(y)\nand\nsin(iy) = (exp(-y) - exp(y))/(2*i) = i*sinh(y)\n\nSubstituting these identities in for cos(iy) and sin(iy), we have:\n\n= (cosh(y))^2 - (i*sinh(y))^2 \n\n= (cosh(y))^2 + (sinh(y))^2\nQED
49770
There is no closed form for the nth derivative of this function.  I symbolically calculated and factored the first 50 derivatives on MathCAD, and there was not a pattern.  Sorry, but you're out of luck....