CALCULUS HELP. please?
918808
wouldnt it be like 3<x<-3 f(x)\nso like x would be 1 so it would be just f if you really want to simply that far done.
982897
the 1st one is simple\n(x+y) dy/dx + y =0\nor xdy/dx + ydy/dx + y =0\nor (xdy/dx + y)  + ydy/dx = 0\nxdy/dx + y = d/dx(xy)\n\nand d(y^2/2)/dx \n\nso d/dx(xy+y^2/2) = 0\n\nor xy+y^2/2 = c\nor 2xy+y^2 = 2c or C\n\nproved\nfor the second one\n\ndy/dx -(1/x)y = xy^2\n\nlet y/x = t\ny = xt\ndy/dx -y/x = t+ x dt/dx -t = x dt/dx\nso\nx dt/dx = - x (xt)^2\nor dt/dx = x^2t^2\ndt/t^2 = x^2 dx\nyou should be able to proceed
889431
1. Find the antiderivative of 2x + 2sin(x).\nIt is f'(x) = x^2 - 2cos(x) + C\nSince f'(0) = 2, then you can plug in 0 for x and set this equal to 2.\n0^2 - 2cos(0) + C = 2\n-2 + C = 2\nC = 4\nf'(x) = x^2 - 2cosx + 4\nNow find this antiderivative.\nf(x) = (1/3)x^3 - 2sinx + 4x + C\nSince f(0) = 2, we have\n(1/3)0^3 - 2sin(0) + 4(0) + C = 2\nC = 2\nSo f(x) = (1/3)x^3 - 2sin(x) + 4x + 2\nTo find f(2) just replace x with 2 and evaluate.\n2.  Find the antiderivative of 7x +4\nIt is f(x) = (7/2)x^2 + 4x + C\nSince it passes through the point (9,5) that means x = 9 and f(x) = 5\n5 = (7/2)9^2 + 4(9) + C\n5 = (7/2)81 + 36 + C\n5 = 567/2 + 36 + C\n5 = 639/2 + C\nC = 5 - 639/2 = -629/2\nThus, f(x) = (7/2)x^2 + 4x - 629/2\nPlug in x = 3 and solve for f(x).
1102373
i wont answer.... log onto www.mathguru.com and discover solutions with explanations.
750296
-2x + h\nIn about 5 minutes you will be taken the limit of the expression as h ---> 0.  Then, you will be told that then it will be the derivative.\n\nAfter that you take 2 - x^2 do the derivative and get -2x.
152408
Use excel to give delta y/delta x on the curve. Look for a point where this value (the derivative) shoots up in value. That's your point.
796927
try the link below...... hope it can put an end to ur integral problems
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
787158
y = xtan(x)\n\nUse product rule:\ndy/dx = x/cos²(x) + tan(x)\n= x/cos²(x) + sin(x)/cos(x) \n= [x + sin(x)·cos(x)] / cos²(x)\n\nUse product and chain rules:\nd²y/dx² = 1/cos²(x) + (-2)·(x/[cos(x)]^3)·(-sin(x)) + 1/cos²(x)\n= 2/cos²(x) + [2·x·sin(x)] / [cos(x)]^3\n= [2·cos(x) + 2·x·sin(x)] / [cos(x)]^3
842257
divide through\n\ny = [x^(3/2)]/3 - 2x^(1/2) -x^(-1/2)\n\nthen take the derivative\n\ndy/dx = [x^(1/2)]/2 - x^(-1/2) + [x^(-3/2)]/2\n\nthen put x^(3/2) in the denominator\n\ndy/dx = [1/x^(3/2)]*{ [x^2]/2 - x + 1/2 }\n\nwe take out the one half to find\n\ndy/dx = [1/2x^(3/2)] * [ x^2 - 2x + 1 ]\n\nwhich is the same as writing \n\ndy/dx = [1/2x^(3/2)] * (x - 1)^2\n\nor a = -1 and b = 2
667880
a) dy/dx = x - 1/x^2\ndy = (x-x^(-2))dx\ny = x^2/2 + 1/x + C\ny(1) = 1^2/2 + 1/1 + C = 2\n1/2 + 1 + C = 2 implies C = 1/2\nAnswer: y = x^2/2 + 1/x + 1/2, x>0\n\nb) The integration is already finished.  Just plug in -1 for x.\ny(-1) = (-1)^2/2 + 1/-1 + C = 1\n-1/2 - 1 + C = 1 implies C = 3/2\nAnswer: x^2/2 + 1/x + 3/2, x<0\n\nc) Just take the derivative of each one.  C1 and C2 are constants, so their derivative is 0.  \n\nd) C1 = 1/2 and C2 = 3/2\n\ne) Follow parts (a) and (b) except plug in 2 for x and set it equal to -1.  Similarly for y(-2) = 2.\nCheck that C1 = -7/2 and C2 = 1/2
927535
#1 \n\nIf f(x)= √(x³ + 5x +121) * (x² +x +11) \n\nthen, f(0)= √(121) * (11) \n= 11 * 11\n= 121\n\nf'(0) = -1/2 (See #3 below at ***)\n\n#2 \nIf y is a differentiable function of x, then the slope of the curve xy^2 -2y + 4y^3 = 6 at the point where y = 1 is what?\n\nWhen y = 1\n\nx(1)^2 - 2(1) + 4(1)^3 = 6 \nie x + 2 = 6\nSo x = 4\nSo y^2 + 2xydy/dx -2dy/dx + 12y^2dy/dx = 0\nSo 1 + 12 dy/dx - 2dy/dx + 12 dy/dx = 0\nThus 22dy/dx = -1\ndy/dx = -1/22\nie slope of tangent at (1, 6) is -1/22\n\n#3\nFind all open intervals on which the following function is decreasing:\nf(x) = x / (x^2 + x -2)\n= x/[(x + 2)(x - 1)]\n= 2/[3(x + 2)] + 1/[3(x - 1)] (by partial fractions)\nDiscontinuities at x = -2 and x = 1\n\nf'(x) = -2/[3(x + 2)²] - 1/[3(x - 1)²]\n\nSo f'(0) = -2/[3(2)²] - 1/[3(-1)²]\n= -2/12 - 1/3\n= -1/2 ***\n\nClearly f'(x) < 0 for all x except x = -2 and x = 1 where it is undefined \n\nie the function is decreasing fo x< -2, -2 < x < 1 and x > 1
121355
well if your taking the derivative, then there should be no h's left over at the end, since they go to zero with the limit.  The derivative of that function is [5(5x+2)+25x]/(5x+2)^2.  I'll leave it to you to simplify.
1166733
dy/dx = {(3x^2-2)(4x-3)^4 - (x^3-2x)(4)(4x-3)^3(4)}/(4x-3)^8\n= {(3x^2-2)(4x-3)^4 - 16(x^3-2x)(4x-3)^3}/(4x-3)^8\n\nUsing the quotient rule.  d/dx f(x)/g(x) = {f'(x)g(x) - g'(x)f(x)}/g(x)^2
850884
using chain rule and product rule\n\ny'= 3x^(2)*tan(4x)\n         +4x^(3)*sec^(2)(4x)\n  = x^2(3tan(4x)+4x*sec^(2)(4x) \n\ni hope that this helps
554917
try Liethold's book...thats our txtbook in school for calculus.
715464
y' = (sinx-cosx) / (sinx+cosx)^2\n\nTo find the tangent at (0,1) you need its coefficient, which is y'(0) = -1.\n\nYour tangent should be y - 1 = -1.(x - 0)\ny = -x + 1