User:Jorgemontes

Límites

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1) $\lim _{x\to 0}{\frac {senx}{2x}}$

${\frac {1}{2}}\lim _{x\to 0}{\frac {senx}{x}}$

${\frac {1}{2}}.1$

${\frac {1}{2}}$ by:Jorge Montes Jorgemontes

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2)$\lim _{x\to -1}{\frac {x^{3}-4x^{2}+x+6}{x+1}}$

$\lim _{x\to -1}{\frac {(x+1)(x^{2}-5x+6)}{x+1}}$

$\lim _{x\to -1}\ {x^{2}-5x+6}$

$\ {(-1)^{2}-5(-1)+6}$

$\ {1+5+6}$

$\ {12}$ by:Jorge Montes Jorgemontes

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3)$\lim _{x\to 3}{\frac {x^{4}-18x^{2}+81}{(x-3)^{2}}}$

$\lim _{x\to 3}{\frac {(x-3)(x^{3}+3x^{2}-9x-27)}{(x-3)^{2}}}$

$\lim _{x\to 3}{\frac {x^{3}+3x^{2}-9x-27}{x-3}}$

$\lim _{x\to 3}{\frac {(x-3)(x^{2}+6x+9)}{x-3}}$

$\lim _{x\to 3}\ {x^{2}+6x+9}$

$\ {(3)^{2}+6(3)+9}$

$\ {9+18+9}$

$\ {36}$ by:Jorge Montes Jorgemontes

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4)$\lim _{h\to 0}{\frac {(x+h)^{2}-x^{2}}{h}}$

$\lim _{h\to 0}{\frac {x^{2}+2hx+h^{2}-x^{2}}{h}}$

$\lim _{h\to 0}{\frac {2hx+h^{2}}{h}}$

$\lim _{h\to 0}{\frac {(h)(2x+h)}{h}}$

$\lim _{h\to 0}\ {(2x+h)}$

$\ {(2x)}$ by:Jorge Montes Jorgemontes

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5) $\lim _{x\to 0}{\frac {tan^{2}x}{senx}}$

$\lim _{x\to 0}{\frac {(sen^{2}x/cos^{2}x)}{senx}}$

$\lim _{x\to 0}{\frac {2senxcosx}{(senx)(cos^{2}x-sen^{2}x)}}$

$2\lim _{x\to 0}{\frac {cosx}{(cos^{2}x-sen^{2}x)}}$

$2\lim _{x\to 0}{\frac {cosx}{cos^{2}x-(1-cos^{2}x)}}$

$2\lim _{x\to 0}{\frac {cosx}{2cos^{2}x-1}}$

${\frac {2\lim _{x\to 0}\ cosx}{2\lim _{x\to 0}\ 2cos^{2}x-1}}$

${\frac {(2)(1)}{(2)(1)-1}}$

$\ 2$ by:Jorge Montes Jorgemontes

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6)${\frac {1}{\sqrt[{2}]{3x}}}$

$\lim _{h\to 0}{\frac {{\frac {1}{\sqrt[{2}]{3x+3h}}}-{\frac {1}{\sqrt[{2}]{3x}}}}{h}}$

$\lim _{h\to 0}{\frac {\frac {{\sqrt {3x}}-{\sqrt {3x+3h}}}{({\sqrt {3x+3h}})({\sqrt {3x}})}}{h}}$

$\lim _{h\to 0}{\frac {{\sqrt {3x}}-{\sqrt {3x+3h}}}{(h)(({\sqrt {3x+3h}})({\sqrt {3x}}))}}$

$\lim _{h\to 0}{\frac {{\sqrt {3x}}-{\sqrt {3x+3h}}}{(h)(({\sqrt {3x+3h}})({\sqrt {3x}}))}}$

$\lim _{h\to 0}{\frac {{\sqrt {3x}}-{\sqrt {3x+3h}}}{({\sqrt {h^{2}}})(({\sqrt {3x+3h}})({\sqrt {3x}}))}}$

$\lim _{h\to 0}{\frac {{\sqrt {3x}}-{\sqrt {3x+3h}}}{({\sqrt {(h^{2})(3x^{2}+9hx)}})}}$

$\lim _{h\to 0}{\frac {{\sqrt {3x}}-{\sqrt {3x+3h}}}{\sqrt {(3x^{2}h^{2}+9h^{3}x)}}}$

$\lim _{h\to 0}{\frac {{\sqrt {3x}}-{\sqrt {3x+3h}}}{\sqrt {3x^{2}h^{2}+9h^{3}x}}}$

$\lim _{h\to 0}{\frac {{\sqrt {3x}}-{\sqrt {3x+3h}}}{\sqrt {(3h^{2}x)(3x+3h)}}}$

Derivadas

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1)$\lim _{h\to 0}{\frac {(2+h)^{2}-4}{h}}$

$\lim _{h\to 0}{\frac {4+4h+h^{2}-4}{h}}$

$\lim _{h\to 0}{\frac {4h+h^{2}}{h}}$

$\lim _{h\to 0}{\frac {h(4+h)}{h}}$

$\lim _{h\to 0}\ {4+h}$

$\ {4+0}$

$\ {4}$ by:Jorge Montes

Jorgemontes

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2)Derivar:$\ F(x)={\frac {(x+1)^{2}}{x-1}}$

$\lim _{h\to 0}{\frac {{\frac {(((x+h)+1)^{2})}{(x+h)-1}}-{\frac {(x+1)^{2}}{x-1}}}{h}}$

$\lim _{h\to 0}{\frac {\frac {hx^{2}+h^{2}x-3h-2hx}{(x+h-1)(x-1)}}{h}}$

$\lim _{h\to 0}{\frac {(h)(x^{2}+hx-3-2x)}{(h)((x+h-1)(x-1))}}$

$\lim _{h\to 0}{\frac {x^{2}+hx-3-2x}{(x+h-1)(x-1)}}$

${\frac {x^{2}-2x-3}{(x-1)^{2}}}$ by: jorge montes Jorgemontes

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3)Hallar la derivada de:

y=x

$\ y'=1$

$\ lim_{x\to 0}{\frac {f(x+h)-F(x)}{h}}$

$\ lim_{x\to 0}{\frac {x+h-x}{h}}$

$\ lim_{x\to 0}{\frac {h}{h}}$

$\ 1$

by: jorge montes and Zorraidorsito Jorgemontes

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4)Derive:${\frac {6}{x^{2}+1}}$

$\lim _{h\to 0}{\frac {{\frac {6}{(x+h)^{2}+1}}-{\frac {6}{x^{2}+1}}}{h}}$

$\lim _{h\to 0}{\frac {\frac {6x^{2}+6-6x^{2}-12hx-6h^{2}-6}{((x+h)^{2}+1)(x^{2}+1)}}{h}}$

$\lim _{h\to 0}{\frac {\frac {-6h^{2}-12hx}{((x+h)^{2}+1)(x^{2}+1)}}{h}}$

$\lim _{h\to 0}{\frac {-6h^{2}-12hx}{(h)(((x+h)^{2}+1)(x^{2}+1))}}$

$\lim _{h\to 0}{\frac {(h)(-6h-12x)}{(h)(((x+h)^{2}+1)(x^{2}+1))}}$

$\lim _{h\to 0}{\frac {-6h-12x}{((x+h)^{2}+1)(x^{2}+1)}}$

${\frac {-6(0)-12x}{((x+(0))^{2}+1)(x^{2}+1)}}$

${\frac {-12x}{(x^{2}+1)^{2}}}$ by: jorge Montes Jorgemontes

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5)Derive por implícita: $\ {y^{2}-x^{2}=1}$

$\ {{\frac {d}{dx}}y^{2}-{\frac {d}{dx}}x^{2}={\frac {d}{dx}}1}$

$\ {{\frac {d}{dx}}2y-{\frac {d}{dx}}2x={\frac {d}{dx}}0}$

$\ {{\frac {d}{dx}}2y-{\frac {d}{dx}}2x=0}$

$\ {{\frac {d}{dx}}(2y-2x)=0}$

$\ {{\frac {d}{dx}}(2y)=2x}$

$\ {{\frac {d}{dx}}={\frac {2x}{2y}}}$

$\ {{\frac {d}{dx}}={\frac {x}{y}}}$

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6)calcular $\ f^{,}(0),f^{,}({\frac {1}{2}}),f^{,}(1),f^{,}(-10)$

$\ {f(x)=2+x-x^{2}}$

$\ {f^{,}(x)=1-2x}$

$\ {f^{,}(0)=1}$

$\ {f^{,}({\frac {1}{2}})=0}$

$\ {f^{,}(1)=-1}$

$\ {f^{,}(-10)=-19}$

Recta Tangente

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Encuentre la ecuación de la recta tangente en el punto propuesto

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1)$\ {x^{3}y+y^{3}x=30}$

Parcial

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1)Se da una tabla de valores para f, g, f´ y g´

/ x / f(x) / g(x) / f´(x) / g´(x) /

/ 1 / 3 / 2 / 4 / 6 /

/ 2 / 1 / 8 / 5 / 7 /

/ 3 / 7 / 2 / 7 / 9 /

Si h(x) = g(f(x)), el valor de h´(1) es igual a:

a)4 b)5 c)36 d)24 e)30

R/

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2)

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3)Un rectángulo tiene dos vértices sobre el eje x y otros dos sobre la parabola $\ {y=12-x^{2}}$ ¿Cuáles son las dimensiones del rectángulo de este tipo con área máxima?

$\ {A=(2x)(y)}$

$\ {A=(2x)(12-x^{2})}$

$\ {A=24x-2x^{3}}$

$\ {A^{,}=24-6x^{2}}$

$\ {0=24-6x^{2}}$

$\ {24=6x^{2}}$

$\ {{\frac {24}{6}}=x^{2}}$

$\ {4=x^{2}}$

$\ {{\sqrt {4}}=x}$

$\ {2=x}$

$\ {y=12-x^{2}}$

$\ {y=12-(2)^{2}}$

$\ {y=12-4}$

$\ {y=8}$

$\ {A=(2(2))(8)}$

$\ {A=(4)(8)}$

$\ {A=32}$

--Jorgemontes 13:07, 19 Jul 2004 (UTC)

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4)