Calculus Mean Value Theorem?
914585
f(x) satisfies the mean value theorem because it is continuous and differentiable on [-2, 1].\nSecant slope = (1 - 4) / (1 + 2) = -1\nDerivative f'(x) = 2x\n2x = -1\nx=-1/2 (this is what c equals to)
492355
Check out Stewart's Calculus(refer to the first link). If you are looking for something  more rigorous refer to Apostol's Calculus(refer to second link).
889651
It's just arctan (x^2).\n\nBy the fundamental theorem of calculus, if F(t) is the antiderivative of f(t) = arctan(t^2), then the integral of f(t) from 0.5 to x will give F(x) - F(0.5). The derivative of this function with respect to x then gives you f(x).
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
517319
Using the definition of the derivative.\nf'(x) = lim(as h->0) of [f(x+h)-f(x)]/h\n\nLet f(x) = mx + b\nf(x+h) = m(x+h) + b\n\nf(x+h) - f(x)\n= m(x+h) + b - (mx+b)\n= mx + mh + b - mx - b\n= mh\n\n[f(x+h)-f(x)]/h \n= mh/h\n= m\nTherefore,\nf'(x) = lim(as h->0) of [f(x+h)-f(x)]/h\n= lim(as h->0) of m\n= m
887032
This is a fundamental theorem of calculus problem.\n\nThe derivative of the integral of t * sqrt (t^3 + 2t) dt from -1 to x is\n\nx * sqrt (x^3 + 2x)
113770
http://mathforum.org/library/drmath/sets/high_calculus.html\n\ngood luck...
553253
http://en.wikipedia.org/wiki/Liskov_substitution_principle click the link to know in detail.
711470
Try here:/\n\nhttp://archives.math.utk.edu/visual.calculus/
1060074
d (ln(x)) / dx = 1/x\n\nThe derivative of the natural log of x with respect to x equals 1 / x
453647
93.16667 =- 3*10.63 \n61.28 to 105.04, so all of the values fit in this range.\n\nThe previous answerers have been wrong about Chebyshev Inequality.  According to that, there should be 1-(1/k)^2 values within k standard deviations of the mean.  So, according to the theorem, there should be 1-(1/3)^2 = 8/9 = 88.89% within 3 SD's of the mean.
320479
Use Rolle's theorem? :-)\n\nRolle's Theorem\nhttp://mathworld.wolfram.com/RollesTheorem.html\nLet f be differentiable on the open interval (a,b) and continuous on the closed interval [a,b]. Then if f(a) = f(b), then there is at least one point c in (a,b) where f'(c) = 0.\n\nGiven: f(x) = g(x)(h(b) - h(a)) - h(x)(g(b) - g(a))\nwhere g and h are differentiable on (a,b)\n{and continuous on [a,b] -- differentiability on (a,b) only implies continuity on (a,b)}\n\nThen f(a) = g(a)(h(b) - h(a)) - (h(a)(g(b) - g(a))\n = g(a) h(b) - g(a) h(a) - g(b) h(a) + g(a) h(a)\n = g(a) h(b) - g(b) h(a)\n\nand f(b) = g(b)(h(b) - h(a)) - h(b)(g(b) - g(a))\n = g(b) h(b) - g(b) h(a) - g(b) h(b) + g(a) h(b)\n = g(a) h(b) - g(b) h(a)\n = f(a)\n\nSo f(a) = f(b) therefore there is a y in (a,b) -- called c in the statement of Rolle's theorem above -- such that f'(y) = 0 = g'(y)(h(b) - h(a)) - h'(y)(g(b) - g(a)) and so at that point y you have g'(y)(h(b) - h(a)) = h'(y)(g(b) - g(a)) as desired.\n\nSee http://cr.yp.to/papers/calculus.pdf
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.
910126
I'm assuming this is with metric spaces D with the metric d and ℝ with the Euclidean metric.\n\nProof.\n\nLet ε > 0. Since both f and g are uniformly continuous, ∃ δ₁, δ₂ > 0 such that for ∀ x, y ∈ D,\n\nd(x, y) < δ₁ ⇒ |f(x) - f(y)| < ε / 2, and\nd(x, y) < δ₂ ⇒ |g(x) - g(y)| < ε / 2.\n\nLet δ = min(δ₁, δ₂). Then d(x, y) < δ implies\n\n|(f+g)(x) - (f+g)(y)|\n= |f(x) + g(x) - f(y) + g(y)|\n= |f(x) - f(y) + g(x) - g(y)|\n≤ |f(x) - f(y)| + |g(x) - g(y)|\n< ε / 2 + ε / 2\n= ε. ∎
920889
if f is differentiable in x+0 then its also continue in x_0 :\n\nf(x) - f(x_0) = f'(x_0)(x-x_0)+ E(x)\n\nlim x->x_0 then left side 0 and right side 0 thus f(x) = f(x_0) for x->x_0\nthus continous in x_0
830837
If ∫f(x)dx = y(x), then f(x) = d y(x) / dx;  So to get the answer, take the derivative of the expression x^2 - 2x + 1.  The value of the integral at limit 1 is  a constant and will not appear in the derivative.
345069
This can be shown for X as either continuous or discrete.  Let mu = E(X).\n\nVar(Y) = E(Y^2) - E(Y)^2\n= E[(X-mu)^2] - E(X-mu)^2\n= E(X^2 - 2Xmu + mu^2) - (E(X)-mu)^2\n= E(X^2) - 2muE(X) + mu^2 - (mu-mu)^2\n= E(X^2) - 2mu^2 + mu^2 - 0\n= E(X^2) - mu^2\n= E(X^2) - E(X)^2\n= Var(X)\n\nIf you need to be more specific than this, the replace E(X) = int(-inf to inf) xf(x)dx and E(X^2) = int(-inf to inf) x^2f(x)dx.  Everything should come out the same way.\n\nRemember that using substitution in the integral, let x = y + mu, the dx = dy, and the limits of integration would remain -inf to inf.