Some calculus questions...?
818391
csc (theta) cot (theta) -3, P(pi/2, 0)\n\n∫[cscθ cotθ - 3]dθ\n= -cscθ - 3θ + c\n\nWhen θ = π/2 expression = 0\nie -cscπ/2 - 3π/2 + c = 0\nSo c = 3π/2 + 1\n\nThus particular integral sought is -cscθ - 3θ + 3π/2 + 1\n\nMean Value Theorem states that in [a, b] there exists c such that \nf'(c) = [f(b) - f(a)]/[b - a] if f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b)\n\nf(x) = x^⅓; a = -1, b = 5\n[f(5) - f(-1)]/[5 - -1]\n= [5^⅓ + 1]/6\n\nf'(x) = 1/[3x^⅔]\n\nSo f'(c) = 1/[3c^⅔]\n\nNow solve 1/[3c^⅔] = [5^⅓ + 1]/6\nSo 3c^⅔ = 6/[5^⅓ + 1]\nie c^⅔ = 2/[5^⅓ + 1]\nc ≈ 0.6340 which is in the interval [-1, 5]\n\nHowever while f(x) is continuous on the closed interval [-1, 5] it is NOT differentiable at 0 which lies on the open interval (-1, 5) so the theorem falls over on this count
1112829
I hope I understood the statement of the problem correctly.\nYou are asking for a set of conditions when an area will be negative.\n\n1. Let f(x)<0 then the area must lie in the positive x domain (x>0).\n2. Let f(x)>0 the area must lie in the negative x domain (x<0).
706213
Wait a minute, did you want to know when the function is zero or when its derivative is zero?  \n\nFor derivative:\ny' = 2x + sin x, right?\n\nIt looks like there's one place where it's 0.  Requires and numerical solution.\n\nWell, it's 0 at x=0, right?\nI assume that x is in radians.  1 radian is about 57 deg.  As x decreases from 0 both parts become more negative.\n\nIt looks like x = 0 is the only zero of the dirivative.
986880
Do you want how to derive them or what the derivatives are?\n\nF(x)=a^x  ---> f'(x) = a^x (ln a)\n\nf(x) = sin x ---> f'(x) = cos x\n         cos x      --->           -sin x\n         tan x       ---->           sec^2 (x)\n         sec x      ---->           sec x tan x\n         csc x      ---->           - csc x cot x\n         cot x       ---->            - csc^2 (x)
1032260
1. if x=f(t) and y=f(t) then\n\nx^4+y^4=6y\n\nd/dt { x^4+y^4=6y}\n\n4x^3 (dx/dt) + 4y^3(dy/dt) =6(dy/dt)\n\nand you have to have the x=f(t) & y=f(t) in order to substitute for the derivatives.\n\n2. if y=f(x) and x=f(t) then\n\nx^4+y^4=6y\n\nd/dt { x^4+y^4=6y}\n\n4x^3 (dx/dt) + 4y^3(dy/dx)(dx/dt) =6(dy/dt)(dx/dt)\n\nthen substitute for the derivatives of the functions y=f(x) & x=f(t)
1101719
Step 1: Find the slope of the line x-2y = 2.  Call it m. \n\nStep 2: Find the derivative f'(x) of the function f(x) = (x-1)/(x+1).\n\nStep 3: Find all values of x where f'(x) = m.\n\nStep 4: For each such value, find its corresponding y-coordinate by setting y=f(x), and then find the equation of the line that goes through that point with slope m.\n\nGood luck!
21911
Try "Calculus" - By Gilbert Strang\nThis is available along with Student study guide &\nInstructor's manual!\n\nhttp://ocw.mit.edu/ans7870/textbooks/Strang/strangtext.htm
118732
good luck man... http://mathforum.org/library/drmath/sets/high_calculus.html
176564
http://www.math.com/\nhttp://mathworld.wolfram.com/\nhttp://www.calculus-help.com/\nhttp://math.about.com/od/calculus/
957759
a. 0 = 7-x^2, x^2=7, x = +/-sqrt(7)\nArea = Integral[-sqrt(7), sqrt(7)] (7-x^2)dx = 7x - (1/3)x^3 |-sqrt(7), sqrt(7) = 2[7sqrt(7) - (1/3)sqrt(7)^3] ~= 24.69 u^2\n\nb. 0 = x^4-8, x^4 = 8, x = +/-8^(1/4)\nArea = (1/5)x^5-8x |-8^(1/4), 8^(1/4) = 2[(1/5)8^(5/4)-8(8)^(1/4)] ~absv= 21.53 u^2
954215
1.\nSince f(0) and f(pi) are both equal to 0 and f is continuous over the interval [0,pi] and differentiable over the interval (0,pi), the conditions of Rolle's theorem are satisfied. Then there are values c, for which f'(c)=0, where f' is the first derivative of f(x) with respect to f.\n\nAll you have to do is find the number of solutions to 0=f'(x)=2cosx + 4cos4x. Using the Pythagorean identity, you can change this into a quartic equation in cosx, so there are at most 4 solutions. To be certain, graph the function on your calculator and count the number of turning points (minima/maxima or extrema) in the interval.\n\n2.\nf'(c)=(f(b)-f(a))/(b-a)\nPlugging in b=5 and a=0, we need to find the value of c satisfying:\nf'(c)=95/5=19\nSince f(x)=x^3 - 6x, f'(x)=3x^2 - 6.\nThen f'(c)=3c^2 - 6=19.\n3c^2=25\nc=5/sqrt{3}
816980
1 is false. The function is continuous.\n2 is false; see above.\n3 is true.  Hence, 4 is false, and so is 5.\nYou are right about the non-differentiability at x=3.  A function cannot be called differentiable if its slope at x- does not approach its slope at x+ as the interval approaches zero.
1128734
I'm going to make a wild guess that you are trying to check an answer you got with the answer in the back of the book. I'm also going to assume that you were supposed to take a derivative of\n(x^3/(a-x))^(1/2).\nIn that case, you got the wrong answer because you didn't do the chain rule. You still need to multiply what you have by the derivative of the thing inside, i.e. the derivative of x^3/(a-x).\n\nIf I'm making a bad assumption and this didn't come from a differentiation problem, please ignore my answer.
934742
uh, find the derivative mathematically and then graph it.\n\nfor example to graph derivative of x^2.\n\nfind the derivative equation f(x)=2x\nand then graph y=2x.\n\nooops, didnt realize it was a picture.\n\nfind the zeroes (a high or low point i.e. tangent is a horizontal line)\nyou should be able to track poitive and negative slopes.  not too hard, you'll figure it out.
113770
http://mathforum.org/library/drmath/sets/high_calculus.html\n\ngood luck...
923293
1. If you pass a vertical line down the graph, and it only passes through the function once, then ya, it is function.  If it passes through more than once, then it is not a function.\n\n2. Isolate x
192062
Maybe this link will help:\nhttp://www.mathinvestigations.com/