·
Cursos Gerais ·
Matemática
Send your question to AI and receive an answer instantly
Recommended for you
11
Ita1976 -1977 - 1978 - Mat - Etapa
Matemática
UMG
3
Teorema de Pitagoras
Matemática
UMG
4
Usp_m_d_funcoes
Matemática
UMG
1
Exercício ENEM
Matemática
UMG
2
Produtos Notáveis- Exercício 1
Matemática
UMG
11
Apostila de Provas Eear - 2013-1
Matemática
UMG
2
Resumo Método da Borboleta
Matemática
UMG
2
Produtos Notáveis- Exercício 1 3
Matemática
UMG
6
Probabilidade
Matemática
UMG
8
Probabilidade e Genética
Matemática
UMG
Preview text
y = 2x^4 - uv\n x^2 e^x\n\ny' = (x^3 e^x)(2x^4) - (x e^x)(2x^4)\n (x^2 e^x)^2\n\ny' = (7x^3 e^x)((2x^4) + (1)(x^1 - 2x^4) - (2x^1) e^x)\n (fx^2 e^x)^2\n\ny' = 8x^3 + 8x^2 e^x - (2x^4 + 2x^1 - x^4)\n (fx^2 e^x)^2\n\ny' = 8x^4 + 8x^3 e^x - 2x^3 - 2 x^2 l e^x\n (fx^3 e^x)^2\n\ny' = 6x^4 + 8x^3 e^x\n (fx^3 e^x)^2 y = 2x - 1\n x - 1\n\ny = (x - 1)(2x + 1)/(x - 1)(2x + 1)\n (x - 2)^2\n\ny' = (x - 1)(2x^2 + 0)/(x - 1)^2(x + 1)(2x + 1)\n\ny' = (x - 1)(2x^0 - x^0)(2x + 1)\n (x - 1)^2\n\ny' = (x - 1)(2.1) - 1(2x + 1)\n (x - 1)^2\n\ny' = (x - 1)(2) - (1)(2x + 1)\n (x - 1)^2\n\ny' = (x - 1)(2 - 2x - 1)\n (x - 1)^2\n\ny' = 2x - 2 - x - 1\n (x - 1)^2\n\ny' = -3\n (x - 1)^2 y = 4x^2 e^x\n\ny' = 4x^2 e^x + 4x^1 e^x\n\ny' = 4x^2(e^x) + (4)(2x^2) e^x\n\ny' = 4x^2 (1) + (8x^1) e^x\n\ny' = 4x^2 e^x + 8x e^x y = 2x^4\n\nx^2 + e^x\n\ny = (x^2 e^x) / (2x^3) - (bx^2 + e^x)' / (2x^2)\n\ny' = (x^2 e^x)*(2.4x^4 - 1) - (2x^2 - 1)(x^2)\n\ny' = (x^2 e^x)(x^3) - (2x^4 + e^x)(2x^3)\n\ny' = (x^2 . 8x^3 e^x)*(2x^3 - (2)2x^4 + 4x^2 e^x)\n\ny' = 8x^5 + 8x^3 e^x - (x^4 + 2x^2)(e^x)\n\ny' = 4x^5 + 8x^3 e^x - 2x^2 e^x\n y = (x^3 + 2x) e^x\n\ny = x + 1\n\ny' = (x^2)(x^1)'(e^x + 2x^2)' e^x\n\ny' = (x^2')'(e^x) + (2x)(e^x) + (2x^2)\ne^x\n\ny' = x^2 e^x + (2x)e^x\n\ny' = x^2 e^x + 4x^2 e^x + 2e^x\n\ny' = x^2 e^x + 4 + 2e^x\n y = x + 1\n\ny = (x^3 + x - 2)(x + 1)' - (x^2 + 2)(x)'\ny = (x^3 + x - 2) / (x^3 + 2) ^ 2\n\ny' = (3x^2 + x)(1)/(3x^2 - 1) - (x^3 + 2)'(x + 1)/(3x^2 - 2)^2\n\ny' = (x^3 + x - 2) * (x)/(x^3 - 1)\n\ny' = (3x^2 + x^3) - 3x^2 - 2x\n\ny' = -2x^3 - 3x - 3\n/ (x^3 + x - 2) ^ 2 g(x) = \\frac{3x-1}{2x+1} \n\ng'(x) = \\frac{(2x+1)(3 \\cdot 1) - (2 \\cdot 1)(3x-1)}{(2x+1)^2} \n\ng'(x) = \\frac{(2x+1)(3 \\cdot 0) - (2)(3x-1)}{(2x+1)^2} \n\ng'(x) = \\frac{(2x+1)(3 \\cdot 1) - (2)(3x-1)}{(2x+1)^2} \n\ng'(x) = \\frac{(2x+1)3 - (2)(3x-1)}{(2x+1)^2} \n\ng'(x) = \\frac{2x+1}{(2x+2)^2} \n\ng'(x) = \\frac{5}{(2x+4)^2} y = \\frac{e^x}{x^2} = u \\rightarrow v = \\frac{V\\cdot M - V_i}{\\sqrt{2}} \n\ny = x^2(e^x) - (x^2) e^x \n\\frac{(fx^2)^2} \n\\ y' = x^2e^x \\cdot - (2x^2-1)e^x \n\\ y' = x^2(e^x \\cdot x^2) - (2x(e^x \\cdot x)) \n\\ y' = x^2e^x - \\frac{X(e^x)}{x^4} y = \\frac{e^x}{1+x} \n\ny = \\frac{y}{V} \n\\rightarrow y = \\frac{V - y - y}{\\sqrt{2}} \n\ny' = (A+x)\\cdot(e^x)' + x^1\\cdot e^x \n\\frac{(A+x)^2}{(A+x)^2} \n\\ y' = \\frac{1+x}{e^x}-e^{-1}+e^x \\cdot \\frac{(1+x)^2}{A+x} \n\\ y' = \\frac{e^x \\cdot x - (1+x)e^x}{(A+x)^2} \n\\ y' = \\frac{\\beta \\cdot x^2}{(A+x)^2}
Send your question to AI and receive an answer instantly
Recommended for you
11
Ita1976 -1977 - 1978 - Mat - Etapa
Matemática
UMG
3
Teorema de Pitagoras
Matemática
UMG
4
Usp_m_d_funcoes
Matemática
UMG
1
Exercício ENEM
Matemática
UMG
2
Produtos Notáveis- Exercício 1
Matemática
UMG
11
Apostila de Provas Eear - 2013-1
Matemática
UMG
2
Resumo Método da Borboleta
Matemática
UMG
2
Produtos Notáveis- Exercício 1 3
Matemática
UMG
6
Probabilidade
Matemática
UMG
8
Probabilidade e Genética
Matemática
UMG
Preview text
y = 2x^4 - uv\n x^2 e^x\n\ny' = (x^3 e^x)(2x^4) - (x e^x)(2x^4)\n (x^2 e^x)^2\n\ny' = (7x^3 e^x)((2x^4) + (1)(x^1 - 2x^4) - (2x^1) e^x)\n (fx^2 e^x)^2\n\ny' = 8x^3 + 8x^2 e^x - (2x^4 + 2x^1 - x^4)\n (fx^2 e^x)^2\n\ny' = 8x^4 + 8x^3 e^x - 2x^3 - 2 x^2 l e^x\n (fx^3 e^x)^2\n\ny' = 6x^4 + 8x^3 e^x\n (fx^3 e^x)^2 y = 2x - 1\n x - 1\n\ny = (x - 1)(2x + 1)/(x - 1)(2x + 1)\n (x - 2)^2\n\ny' = (x - 1)(2x^2 + 0)/(x - 1)^2(x + 1)(2x + 1)\n\ny' = (x - 1)(2x^0 - x^0)(2x + 1)\n (x - 1)^2\n\ny' = (x - 1)(2.1) - 1(2x + 1)\n (x - 1)^2\n\ny' = (x - 1)(2) - (1)(2x + 1)\n (x - 1)^2\n\ny' = (x - 1)(2 - 2x - 1)\n (x - 1)^2\n\ny' = 2x - 2 - x - 1\n (x - 1)^2\n\ny' = -3\n (x - 1)^2 y = 4x^2 e^x\n\ny' = 4x^2 e^x + 4x^1 e^x\n\ny' = 4x^2(e^x) + (4)(2x^2) e^x\n\ny' = 4x^2 (1) + (8x^1) e^x\n\ny' = 4x^2 e^x + 8x e^x y = 2x^4\n\nx^2 + e^x\n\ny = (x^2 e^x) / (2x^3) - (bx^2 + e^x)' / (2x^2)\n\ny' = (x^2 e^x)*(2.4x^4 - 1) - (2x^2 - 1)(x^2)\n\ny' = (x^2 e^x)(x^3) - (2x^4 + e^x)(2x^3)\n\ny' = (x^2 . 8x^3 e^x)*(2x^3 - (2)2x^4 + 4x^2 e^x)\n\ny' = 8x^5 + 8x^3 e^x - (x^4 + 2x^2)(e^x)\n\ny' = 4x^5 + 8x^3 e^x - 2x^2 e^x\n y = (x^3 + 2x) e^x\n\ny = x + 1\n\ny' = (x^2)(x^1)'(e^x + 2x^2)' e^x\n\ny' = (x^2')'(e^x) + (2x)(e^x) + (2x^2)\ne^x\n\ny' = x^2 e^x + (2x)e^x\n\ny' = x^2 e^x + 4x^2 e^x + 2e^x\n\ny' = x^2 e^x + 4 + 2e^x\n y = x + 1\n\ny = (x^3 + x - 2)(x + 1)' - (x^2 + 2)(x)'\ny = (x^3 + x - 2) / (x^3 + 2) ^ 2\n\ny' = (3x^2 + x)(1)/(3x^2 - 1) - (x^3 + 2)'(x + 1)/(3x^2 - 2)^2\n\ny' = (x^3 + x - 2) * (x)/(x^3 - 1)\n\ny' = (3x^2 + x^3) - 3x^2 - 2x\n\ny' = -2x^3 - 3x - 3\n/ (x^3 + x - 2) ^ 2 g(x) = \\frac{3x-1}{2x+1} \n\ng'(x) = \\frac{(2x+1)(3 \\cdot 1) - (2 \\cdot 1)(3x-1)}{(2x+1)^2} \n\ng'(x) = \\frac{(2x+1)(3 \\cdot 0) - (2)(3x-1)}{(2x+1)^2} \n\ng'(x) = \\frac{(2x+1)(3 \\cdot 1) - (2)(3x-1)}{(2x+1)^2} \n\ng'(x) = \\frac{(2x+1)3 - (2)(3x-1)}{(2x+1)^2} \n\ng'(x) = \\frac{2x+1}{(2x+2)^2} \n\ng'(x) = \\frac{5}{(2x+4)^2} y = \\frac{e^x}{x^2} = u \\rightarrow v = \\frac{V\\cdot M - V_i}{\\sqrt{2}} \n\ny = x^2(e^x) - (x^2) e^x \n\\frac{(fx^2)^2} \n\\ y' = x^2e^x \\cdot - (2x^2-1)e^x \n\\ y' = x^2(e^x \\cdot x^2) - (2x(e^x \\cdot x)) \n\\ y' = x^2e^x - \\frac{X(e^x)}{x^4} y = \\frac{e^x}{1+x} \n\ny = \\frac{y}{V} \n\\rightarrow y = \\frac{V - y - y}{\\sqrt{2}} \n\ny' = (A+x)\\cdot(e^x)' + x^1\\cdot e^x \n\\frac{(A+x)^2}{(A+x)^2} \n\\ y' = \\frac{1+x}{e^x}-e^{-1}+e^x \\cdot \\frac{(1+x)^2}{A+x} \n\\ y' = \\frac{e^x \\cdot x - (1+x)e^x}{(A+x)^2} \n\\ y' = \\frac{\\beta \\cdot x^2}{(A+x)^2}