I need help on what the antiderivatives are for these 2 problems?
918747
For #1, there's no way you can solve that directly.  Note that it can be split up into three fractions, though.\n\n(x^2 + x + 1)/x = x^2/x + x/x + 1/x\n\nNow, we can actually reduce that, to\n\nx + 1 + 1/x\n\nWe solve for the antiderivative using the reverse power rule for derivatives.  That is, antiderivative for x^n is [x^(n+1)]/(n+1).  The exception is 1/x, where the antiderivative is ln|x|.\n\nSo the antiderivative of x+1+1/x is\n\n(x^2) / 2 + x + ln|x|\n\nBut we want the general antiderivative, so we add a constant C.\n\n(x^2)/2 + x + ln|x| + C
1139073
Differential Equations
891699
Si? Are you referring to the sine integral? If so, you need only remember two things:\nDifferentiation is the inverse of integration\nd/dx f(g(x)) = f'(g(x)) * g'(x)\n\nSo:\n\nd/dx (Si(x²)) = sin (x²)/(x²) * 2x
1122457
1st question:\ny = x+10\ndy = dx\n\nint((y-6)e^(y-10)/y^7)dy = (e^-10)int(e^y/y^6)dy - 6(e^-10)int(e^y/y^7)dy\n\nTo the side (for integration by parts for the first integral on the right side):\nu = 1/y^6, du = -6/y^7dy\ndv = e^ydy, v = e^y\n\n= e^(-10)((e^y/y^6) - int(e^y)(-6/y^7)dy) - 6(e^-10)int(e^y)/(y^7)dy +c\n= e^-10(e^y/y^6) + 6e^-10int(e^y/y^7)dy - 6e^-10int(e^y/y^7)dy + c\n= e^(y-10)/y^6 + c\n= e^x/(x+10) + c\n\nSecond question:\nDifferential on both sides\n\nyx^(y-1)dx + x^y/ln(y)dy + xy^(x-1)dy + y^x/ln(x)dx = 0\ndx(yx^(y-1) + y^x/ln(x)) + dy(x^y/ln(y) + xy^(x-1)dy) = 0\ndy(x^y/ln(y) + xy^(x-1)) = -dx(yx^(y-1) + y^x/ln(x))\ndy/dx = -(yx^(y-1) + y^x/ln(x))/(x^y/ln(y) + xy^(x-1))\n\nYou really ought to ask these as separate questions though.
99672
honestly....good luck\n\ntry www.mathworld.com\n\ndo a search for differential equations, i think that would be your best bet...but that can get confusing...good luck
706993
Try reviewing your class notes and looking through your text book. Im sure that the answer you need is there. It might take a bit of thought to figure out how to do these problems! Take the time to figure it out, it will be worth it in the long run!
981618
I guess 1) just wants you to write: Want dA/dt, have da/dt and db/dt.\n(From now on I'll use A' to mean dA/dt etc).\n2)\nA=π*a*b, so using the product rule, A' = πa'b + πab'. So A' = 7πb - 11πa.\n3) Just substitute in: A' = 7*π*50 - 11*π*30 = 20π, so thats the rate it is increasing.\n4) Again substitute in, A' = 7*π*20 - 11*π*40 = -300π, so now its decreasing at that rate.
461412
1) assuming it's inversely proportional then you have multiplied by m+2/m so it's dm/(m+2)\n2)(m+2)d/m
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
1138866
I'll lead you in the most important part, but it is up to you to complete it.\n\nDerivative of x . y = x'.y + x.y'\n\nThen, you will have:\n\nf'(x) = (2x . ln x^2) + x^2 (ln x^2)'\n\nContinue from here, the only problem that you may have is finding the derivative of ln y , which you can see in the link below.
919338
Well, I see you've used the correct order of operations this time. That's good - it shows that you're improving.\n\n#1: Apply the product and chain rules to find f'(x), then evaluate at 0:\n\nf'(x) = d(√(x³+5x+121))/dx (x²+x+11) + (2x+1)√(x³+5x+121)\n\nf'(x) = 1/(2√(x³+5x+121)) * d(x³+5x+121)/dx * (x²+x+11) + (2x+1)√(x³+5x+121)\n\nf'(x) = 1/(2√(x³+5x+121)) * (3x²+5) * (x²+x+11) + (2x+1)√(x³+5x+121)\n\nNow evaluating :\n\nf'(0) = 1/(2√121)*5*11 + √121 = 5/2 + 11 = 27/2\n\n#2: First, evaluate the function at y=1 to get the value of x:\n\nxy^2 -2y + 4y^3 = 6\nx - 1 + 4 = 6\nx = 3\n\nNow, implicitly differentiate both sides:\n\nd(xy^2 -2y + 4y^3)/dx = d(6)/dx\ny² + 2xy dy/dx - 2 dy/dx + 12y² dy/dx = 0\n\nSolve for dy/dx:\n\ny² + (2xy - 2 + 12y²) dy/dx = 0\n(2xy - 2 + 12y²) dy/dx = -y²\ndy/dx = -y²/(2xy - 2 + 12y²)\n\nNow evaluate at (3, 1):\n\ndy/dx = -1/(6 - 2 + 12) = -1/16\n\n#3: First, find the derivative:\n\nf'(x) = (x²+x-2 - (2x+1)x)/(x²+x-2)² (using the quotient rule)\n\nSimplifying:\n\nf'(x) = (x²+x-2 - 2x² - x)/(x²+x-2)²\nf'(x) = (-x²-2)/(x²+x-2)²\n\nNow we must find where f'(x) < 0\n\n(-x²-2)/(x²+x-2)² < 0\n-x²-2 < 0 (we know (x²+x-2)² is nonnegative, and if the expression on the left is defined, it is also nonzero. Thus it is positive and we can multiply by it without changing the inequality)\n-x² < 2\n\nNow, since x is a real number, this expression will always be true. Thus, if f'(x) is even defined, x is decreasing. Now f'(x) is defined provied that (x²+x-2)²≠0. So:\n\n(x²+x-2)²≠0\n(x-1)²(x+2)²≠0\nx≠1 and x≠-2\n\nWhich makes sense, since those are vertical asymptotes of the original function. Thus, the function is decreasing on:\n\n(-∞, -2) ∪ (-2, 1) ∪ (1, ∞)
581284
1) d. all\n2) b. decreases\n3) a. 1.3 x 10^-2
677039
(1) { -2x-y-z=0\n(2) { -x-2y-z=0\n(3) { -x-y-2z=0\n\nLet's look at (2):\n-x-2y-z=0 => x=-2y-z\nNow at (3):\n-x-y-2z=0 => x=-y-2z\nNow we can see that\n-2y-z=-y-2z => *** y=z ***\n\nNow let's put these values into (1):\n-2x-y-z=0 => -2x-y-y=0 => -2x-2y=0 => 2x=-2y => *** x=-y ***\n\nFor now, we know that x=-y=-z\nLet's put all of what we know into (2), and express everything using 'x':\n-x-2y-z=0 => -x-(-2x)-(-x)=0 => -x+2x+x=0 => 2x=0 => *** x=0 ***\n\nThus, x=0 => y=-0=0 => z=0\nSo the only solution is (0,0,0).\n\nGood luck!
1015799
Use the second derivative test, not sure exactly what it is still but it should be easy to find on the internet.
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
873172
u can get them with the help of second and higher derivatives.\ninflection point is there where the second derivative is zero.
421380
from first equation put the value of b in second and then form a second degree differential equation in "a"...good luck!