function rule d(c(x))?
921998
is this for math?\nif so:\nTake the function c(x) and put it in for every x in the function d(x).\nSo, if \nc(x) = 2\nand \nd(x) = 3x^2 + 5x -1\nthe new equation would look like\nd(c(x)) = 3(2)^2 + 5(2) -1\nfor a final answer of:\n21
911904
using the rule\n\ndf/dx = df/dy * dy/dx and y = 7x +(49x^2-16)^1/2\n\nwe get df/dy = 1/y = 1/(7x+49x^2-16)^1/2 ...1\n\ndy/dx = 7+ 1/2(49x^2-16)^(-1/2) d/dx(49x^2-16)\n= 7+ 1/(2*49x^2-16)^(1/2) * 98 x ...2\n\nusing d/dx(y^n) = n y^(n-1) dy/dx where y = 49x^2-16\n\nfrom 1and 2 by multiplying\ndf/dx =  (7+ 98x/(2*49x^2-16)^(1/2))/(7x+49x^2-16)^1/2 ...
546449
define f(x)=d(x,A)/[d(x,A)+d(x,B)].\nThis is a continuous function f:X->[0,1] (use triangle inequality to show x->d(x,A) is continuous for a closed set A). Also f=0 on A and f=1 on B. Let C=f^(-1)[0,1/3) and D=f^(-1)(2/3,1]. Then C and D are open because inverse images of open sets under continuous functions are open.\n\nNote to Adolfo Boy: It is possible for the distance between two disjoint closed sets to be 0. Consider the x-axis and the y=1/x curve. Your argument does work for compact sets, though. The point (a,b) in your argument need not exist if the infimum is not attained.
821550
if f(x) = ∫[t = x to 0] 1/ (1 + t + t²) dt\n\nf'(x) = -1/ (1 + x + x²)\n\nf''(x) = (1 + 2x)/(1 + x + x²)²\n\nFor concave UP f''(x) > 0\n\nSo 1 + 2x > 0 (as (1 + x + x²)² > 0 for all x)\n\nie x < -1/2
737185
How about something simpler, like\n\nf(x) = 0 when x<1/2, 1 when x>=1/2\n\nThen the limit from the left is 0, the limit from the right is 1, so the regular limit doesn't exist.
304526
let a-x=u\nso, -dx=du\n\nthen ∫1/(a-x)^2dx= \n-∫1/[(u)^2]du=\n-∫u^(-2)du=\n\n-[u^(-2+1)]/(-2+1)=\n\n-u^(-1)/-1=\nu^(-1)=\n\n1/u + C=\n\n1/(a-x) + C
836547
f(x)=rtx+C\nf(1)=1+C=2\nso C=1\nf(1/4)=rt(1/4)+1=3/2
791940
In order to a od function the function have to obay:\nf(x) = - f (-x) <=>\n-3x^9 + 4x^5 = -(-3(-x)^9 + 4(-x)^5) <=>\n-3x^9 + 4x^5 = 3(-x)^9 - 4(-x)^5 <=>\n-3x^9 + 4x^5 = -3x^9 + 4x^5 ( True)\nthe function is od.
803550
With respect to what?\n\nd/dx xy = y + x dy/dx\n\nd/dy xy = x + y dx/dy\n\nd/dt xy = x dy/dt + y dx/dt\n\nIt's just the Product Rule.
797614
The only problem in your derivation is that you are evaluating f(0, 0) as 0/0. This is not the case, the second part of the definition is that f(x, y) = 0 when (x, y) = (0, 0). Nonetheless, your result is correct: f_x(0, 0) = [h→0]lim (0/h² - 0)/h = 0. Similarly with f_y(0, 0).\n\nThis makes sense, because the restriction of f(x, y) to the x-axis is a constant function (i.e. f(x, 0) = 0 for all x), so of course the partial derivative of it w.r.t. x is constant. Similarly, if you restrict f(x, y) to the y-axis, you also get a constant function.\n\nNote that f(x, y) is not differentiable in the 2-variable sense at (0, 0) (there is no tangent plane at that point), so the multivariate chain rule will NOT hold, and the directional derivatives won't be related to the partial derivatives. This helps to highlight an important difference between the partial and total derivates - it is possible for all partial derivatives to be defined and for the function to still not be differentiable at a point.
914435
x > 5\n\n- - - - - -s-
726082
Yes since ∫f(x) dx = ∫x dx\n= ½x² + c where c is ANY constant.
1033116
A X = D\nX = A^(-1) D. (A inverse * D)\n Do it ur self.
638682
Domain ~ x values\nRange ~ y values\n\nYou have to work the function and figure them out.
1164142
F=O \n\nbut if E=MC2 then F=ND3\n\nthen F=ND3, \nsince ND is just [(M+1)(C+1)] \nthen \nF=({[(M+1)(C+1)]}3)\n\n\nOr you could say that\nF=E+1
861527
Stop, stop, stop. If it was f(x)=2logx, f(x) cannot be = to (1/log2)*logx, as  1/log2 is not = 2.\nDon’t hurry, put your question correctly with necessary details and don’t save on ()-s in formulas.
761744
use the product rule, with the first term as t and the second as (4-t)^1/2\nproduct rule says that the derivative is second term times the derivative of the first term plus first term times the derivative of the second term.  Don't forget to use the chain rule along the way.\n\nThat is\nxy would have a derivate of yx' =xy'\n\nSo here\n[(4-t)^1/2](1) + (t){[1/2(4-t)^ -1/2][-1]\n(4-t)^1/2 + -t/2(4-t)^ -1/2