what is the first dirivative of:  xsiny=ycosx  ?
957209
remember to use the product rule and the chain rule on this.\n\n(dx/dx)(sinY)+xcosy(dy/dx)=-ysinx(dx/dx)+(dy/dx)cosx\n\ndx/dx = 1 \n\nnow  solve algebraicially for dy/dx\n\n1) isolate all the terms with dy/dx\nxcosY(dy/dx)-cosx(dy/dx)=-ysinx-sinY\n\n2) factor out (dy/dx)\ndy/dx(xcosY-cosx)=-ysinx-sinY\n\n3) divide\ndy/dx  = (-ysinx-siny)/(xcosy-cosx)
1048691
(r*x)*(-sin(x))+(cos(x))*(r)+\n(0)*(lnrsin(x))+(r)*(cot(x)) \n\nfor the cot(x) you end up for the derivative of lnrsin(x) is (1/sin(x)) *cos(x)=cot(x)\n\nsorry it cut off on the top. hopefully this works
362083
Please use the search function. This question was asked 5 days ago:\n\nA proper polar coordinate system (one that assigns each point in the plane a unique coordinate) will always leave out a ray of points. For example, the standard polar system that identifies the open set (0, oo) X (0, 2Pi) with R^2 - {0} by the function f(r, t) = (rcos(t), rsin(t)) ( = (x, y) ) does not correspond to any points on the positive x-axis and necessarily leaves out the origin (there is no way to include the origin uniquely). This is usually glossed over in a basic calculus or physics course as one actually uses an atlas of polar coordinate charts (this atlas never includes the origin, but at least covers the plane minus the origin); you use whichever chart is convenient for the problem at hand. Some definitions will disregard the need for a homeomorphism and include the positive x-axis by considering a domain that is not open, (0, oo)X[0, 2Pi)
861275
x(y-1) = 2y + 3\n\nxy - x = 2y + 3\nxy - 2y = x + 3\n(x - 2) = (x + 3) / y\n y = (x + 3) / (x -2)
972809
Consider the left-hand side(LHS) [cot(x) + 1] / [cot(x) – 1]. We want to prove that this is equal to the right-hand side (RHS), that is, [1 + tan(x)] / [1 – tan(x)]. Thus, we have the following:\n\nLHS = [cot(x) + 1] / [cot(x) + 1]\n        =  { [cot(x) + 1] / [cot(x) – 1] }* { tan(x) / tan(x) }.\n\nNOTE: We multiply tan(x) to cot(x) + 1 and the other tan(x) is multiplied to cot(x) – 1.\n\n        =  { tan(x) [cot(x) + 1] } / { tan(x) [cot(x) – 1] }\n\n        =  { tan(x) cot(x) + tan(x)] } / { tan(x) cot(x) – tan(x)] }, by distributive property of multiplication over addition (subtraction)\n\n        =  { 1 + tan(x)] } / { 1 – tan(x)] }, since tan(x) cot(x) = 1 by reciprocal identity.\n\n        =  RHS.\n\nIndeed, [cot(x) + 1] / [cot(x) – 1] = [1 + tan(x)] / [1 – tan(x)]
329632
{x: x inequal to 2}\nor\n{x: x=/= 2}
433823
Revolving about the x-axis yields a cone whose tip is at the origin and opens in the positive x direction.  Taking discrete x slices (dx) yields a disk whose volume is given by pi*(kx)^2 dx.  Integrating from 0 to h gives us (where "S" is used as the integration symbol):\n\nV = S pi*(kx)^2 dx\n= pi * k^2 S x^2 dx\n= (pi * k^2) * (x^3)/3\n\nEvaluating from 0 to h gives us\n= (pi * k^2) * (h^3)/3 - (pi * k^2) * (0^3)/3\n= (pi * k^2 * h^3)/3
308163
x can be any value.\n\ntan x = sin x / cos x\n\ncot x = cos x / sin x\n\nso tan x * cot x = sin x/cos x * cos x/sin x\n\nsin x's cross out each other\n\ncos x's cross out each other\n\nso you're left with 1.
206636
for (i=1,t=1,cosx=1;i<=n;++i)\n{\n  t=(-1)*(t*x*x)/((2*i)*(2*i-1);\n  cosx += t;\n}
80323
No. xy = k is an hyperbola in 2D. xy = kz is an hyperboloid on 3D.
700222
It is always x and then y. thew way i remember it is because it goes in alphabetical order. hope it helps!
405955
If you are reflecting across the y axis, x becomes -x.\n\nIf you are reflecting across the x axis, y becomes -y.\n\nIf you are reflecting about the origin, both x and y change sign.\n\nIs that what you are looking for?
757845
x  y\n0 0\n1 -1/5\n-5 1
668877
it's (xy) / 2 - 1 * 3 + 1 - 2 which is (3xy) / 2 - 2 because remember the order of operations
764433
Finding max's and mins is done thru the first derivitive and setting it to 0, so\n\nf(x)= sinx - cosx\nf1(x)= cosx + sinx\n0=cosx + sinx\n-cosx = sin x\nthis occurs when x= (3pi)/4\nmax is root 2, as was stated.\n\nThe answer that used to be directly above is wrong.
949562
Use csx =1/sin and  1-cos^2=sin^2  and cot=cos/sin\n Ready...Go\ncscx - (cosx)(cotx) =  1/sin - cos^2/sin = (1-cos^2)/sin=sin^2/sin =sin\n       ole, ole , infree
928762
Your will have Yx (Y to the power x)