Lista Calculo

Lista de c\\u00e1lculo\n1) a) f(x) = x^2 - 2x + 1\nf(x0) = x0^2 - 2x0 + 1\nf(x0 + \\u0394x) = (x0 + \\u0394x)^2 - 2(x0 + \\u0394x) + 1\nf(x0) = lim (x0 + \\u0394x) -> x0 [ (\\u2026 ) - (\\u2026) ] =\n\\n\\n\\n\ng(x) = lim_{\\Delta x\\to 0} [\\text{...}]\n\\n\\n\\n\ng'(x) = (\\text{...})\n\\n\\n\\n\\n\\n\\n\\n d) f(1) = x^2 - 2x .\nf(x) = lim (y0 + \\u0394y) -> (y0 + \\u0394y)\\u2026\nf(x) = lim (x0 + \\u0394x) -> (x0 + \\u0394x)\\u2026\n\\n\\n\\n\\nx(y) = \\lim g(x0) = 2\\to f(x0 + \\u0394x)\n\\n\\n\\n\\n\\n\\n a) f(2) = 2x^2 - 2x . f(3) = \\theta\\n\\n\\n\\n\\n\\n\\n\\n\\n11)a) y = 4x + 5\ny' = 4\n\\n\\n\\n\\nb) y = -x + 3\ny' = -1\n\\n\\n\\nc) y = \\frac{1}{2} x^2\ny' = 1\n\\n\\nd) y = x^2 + 4x + 5\ny' = 2x + 4\n\\n\\ne) y = \\frac{1}{2} x^2 + 5x + 7\ny' = -3x + 5\n\\n\\nf) y = 0.1x^2 - 9x\ny' = 0.4x - 4 k) y' = \\sqrt{x}\ny' = \\frac{1}{4 \\sqrt{x^3}}\n1) y' = 9 \\sqrt{x}\ny' = \\frac{1}{9 \\sqrt{x^3}}\nml) y' = 3 \\sqrt{x}\ny' = \\frac{1}{3 \\sqrt{x^2}}\n\nn) y' = \\frac{1}{6 \\sqrt{x^5}}\n0) y' = -\\frac{1}{x}\n\nP) y' = 6 x^3\ny' = -\\frac{15}{x^4}\n\ny' = 1 \\cdot (1-x)^{-2}\n\ny' = \\frac{1}{(1-x)^2}\\ 1) a) f(x) = (2x-2)^3\nf'(x) = 3(2x-2)^2(2) = 6(2x-2)^2\n\nb) g(x) = 3(9x-4)^4\ng'(x) = 3\\cdot 4(9x-4)^3(9)\n\nc) f(t) = (9t+12)^{\\frac{1}{3}}\nf'(t) = \\frac{2}{3}(9t+12)^{\\frac{2}{3}}(9)\n\n\\dot{d} y = \\sqrt{3x^2 - ux}\ny' = \\frac{1}{3}(3x^2 - 4x)^\\frac{3}{2}\n\ny' = g(3x^2 + ux)\n\ne) f(1) = (12 + 1x^2)^{-\\frac{1}{2}}\nf'(x) = \\frac{1}{2}(25 + x)^{-\\frac{3}{2}}(2x)\n\n2) f(x) = \\sqrt{4x^2 - 7}\nf'(x) = \\sqrt{16 - 7} = 1\\n3\n\nf'(x) = \\frac{1}{2}(4x-2)^{-\\frac{1}{2}}(8x)\n\ny'(1) = 8(4 - 2)\n 2) P(1)=\\frac{1}{3}&(P(2,3))\n\nf(x) = 3 = \\frac{2}{3}(8-1)\ny' = \\frac{7x-16 - 13}{3}\n\ny = \\frac{7}{3}x - \\frac{16}{3}\n\nP = (2,\\frac{2}{2 - 1})\n\ny' = -\\frac{15}{(6,8)} - \\frac{3}{3}\n