Calculus Word Problem?
534670
Jimvalentinojr had the derivative right but didn't do enough algebra with it.  The x² he said was in the denominator was in the denominator of a fraction that was part of the denominator of another fraction, so the thing to do is simplify the complex fraction.  We had the derivative as\n\nΘ' = (1/x²)(-(b+a)/[1+(b+a)²/x²]) + (1/x²)(a/[1+a²/x²])\n\nso we multiply top and bottom of the 2 complex fractions by x² and get\n\nΘ' = (1/x²)[-x²(b+a)/(x²+(b+a)²)] + (1/x²)[ax²/(x²+a²)]\n\nand in each term x² cancels out, leaving\n\nΘ' = -(b+a)/[x²+(b+a)²] + a/(x²+a²)\n\nnow setting Θ' = 0,\n\na/(x²+a²) = (b+a)/[x²+(b+a)²], and cross multiplying (denominators never 0)\n\n(b+a)x² + (b+a)a² = ax² + a(b+a)²\n(b+a-a)x² = a(b+a)² - a²(b+a)\nbx² = ab² + 2a²b + a^3 - a²b - a^3\nbx² = ab² + a²b\nx² = ab + a²\nx = sqrt(ab + a²), and of course we ignore the negative root.
806655
dx=2xlny-2z\ndy=x^2*(1/y)-2z\ndz= x^2lny-2
953139
f'(x)=(4/3)x^(1/3) +(4/3)x^(-2/3)
1132233
this question will require calcs using logarithm.\nhence will take time
904469
Let u = e^x + 1.  Then du= ...\n\nHope that helps.
364905
Use Charles' Law\n\nLet V initial = V\nLet T initial = T\n\nLet V final = V'\nLet T final = T'\n\n\nV/T = V'/T'\n\nSolve for V' you might have to convert from mL to L though.
1121843
the given function is not differentiable at x=0\n\ntherefore,   f(x) = x   for x>0        dy/dx = 1\n                           f(x) = -x  for x<0        dy/dx = -1
782468
1.ac\n\n2.-3r\n\n3.3m/-2r\n\n4.(x-4)\n\n5.(a-b)(b-c)
1033142
This is a very standard problem. Since it is impossibe to type this in text, I provide the following link to the solution from the Math World Web site:\nhttp://mathworld.wolfram.com/LegendreDifferentialEquation.html
713982
use logarthmic differentiation\n\ny = x^(1/x)\n\nln(y) = (1/x)* ln(x)\n\ndiff w.r.t x\n\n(1/y) (dy/dx) = (1/x)*(1/x) +ln(x)*(-1/x^2)\n\n(dy/dx) = { y/(x^2) }  [ 1 - ln(x) ]\n\nsubstitute for y
1041686
That's easy:   f'(x)=\n\n2*(x)^(3/2) - 3 - (x)^(1/2)\n--------------------------------\n6*(x)^(5/2)
986088
a) W = mg = 30*9.90662 = 294.1986 N\nb) Ft = 294.1986sinθ = 155.0897 N\nc) cosθ = 250/294.1986, θ  = 31.81376°\nd) a = 155.0897/30 = 5.19657 m/sec^2
347066
ok, i think one will be where you have to solve a word problem dealing with alegebra. One with number sentences that have a letter on one side of the problem and letters on both sides of the problem. i could be wrong, but i'm trying to help.
57176
http://www.calculus.org/
780025
x ≤ 5\n\n-- - - - - - -s-
921398
I just taught this section today to my calc students.  You start by finding the first and second derivative.  I recommend looking in your book for an example.  Your book probably has something like this towards the front of the section.  It would be too long winded to write everything down here.  If you don't have your book on you, the web page below has problems that you can do and then check your answer.\n\nhttp://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/Graphing.html
500014
a. (x+9)(x-9)\nb.(t-2)(t+2)\nc. (b-12)(b+12)