IvanH

1) $\lim _{x\to 9}{\frac {x^{2}-81}{{\sqrt {x}}-3}}$ =$\lim _{x\to 9}{\frac {(x-9)(x+9)}{{\sqrt {x}}-3}}\cdot {\frac {{\sqrt {x}}+3}{{\sqrt {x}}+3}}$ =$\lim _{x\to 9}{\frac {(x-9)(x+9)({\sqrt {x}}+3)}{x-9}}$ =$\lim _{x\to 9}\ {(x+9)({\sqrt {x}}+3)}$ =$\ {((9)+9)({\sqrt {9}}+3)}$ =$\ {(18)(6)}$ =$\ {108}$

2) $\lim _{x\to 0}{\frac {x}{{\sqrt {1+3x}}-1}}$ =$\lim _{x\to 0}{\frac {x}{{\sqrt {1+3x}}-1}}\cdot {\frac {{\sqrt {1+3x}}+1}{{\sqrt {1+3x}}+1}}$ =$\lim _{x\to 0}{\frac {(x)({\sqrt {1+3x}}+1)}{(1+3x)-1}}$ =$\lim _{x\to 0}{\frac {(x)({\sqrt {1+3x}}+1)}{3x}}$ =$\lim _{x\to 0}{\frac {{\sqrt {1+3x}}+1}{3}}$ =${\frac {{\sqrt {1+3(0)}}+1}{3}}$ =${\frac {{\sqrt {1}}+1}{3}}$ =${\frac {2}{3}}$

3) $\ f(x)={x^{2}-10x+100}$ =$\ f'(x)={(2)x^{2-1}-10(1)x^{1-1}+0}$ =$\ f'(x)={2x-10}$

4) Failed to parse (syntax error): {\displaystyle \ f(r)= /frac {4πr^3}{3}} =Failed to parse (syntax error): {\displaystyle \ f'(r)= /frac {(3)4πr^{3-1}}{3}} =Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikibooks.org/v1/":): {\displaystyle \ f'(r)= {4πr^2}}

5) $\ F(x)={(16x)^{3}}$ =$\ F(x)={(16^{3})(x^{3})}$ =$\ F(x)={4096x^{3}}$ =$\ F'(x)={(3)(4096x^{3-1})}$ =$\ F'(x)={12288x^{2}}$

6) $\ Y(t)={6t^{-}9}$ =$\ Y'(t)={(-9)(6t^{-9-1})}$ =$\ Y'(t)={-54t^{-}10}$

7) $\ g(x)={x^{2}+{\frac {1}{x^{2}}}}$ =Failed to parse (syntax error): {\displaystyle \ g'(x)= {(2)(x^{2-1}) + \frac {(0)(x^2)-(1)(2)(x^{2-1})}{(x^2)^2}} =$\ g'(x)={2x}+{\frac {0-2x}{x^{4}}}$ =$\ g'(x)={2x}-{\frac {2}{x^{3}}}$

8) $\ h(x)={\frac {x+2}{x-1}}$ =$\ h'(x)={\frac {(1)(x^{1-1})(x-1)-(x+2)((1)(x^{1-1})}{(x-1)^{2}}}$ =$\ h'(x)={\frac {(x-1)-(x+2)}{(x-1)^{2}}}$ =$\ h'(x)={\frac {-3}{(x-1)^{2}}}$

9) $\ G(s)={(s^{2}+s+1)}{(s^{2}+2)}$ =$\ G'(s)={((2)(s^{2-1})+(1)(s^{1-1}))(s^{2}+2)}+{(s^{2}+s+1)(2)(s^{2-1})}$ =$\ G'(s)={(2s+1)(s^{2}+2)}+{(s^{2}+s+1)(2s)}$ =$\ G'(s)={(2s^{3}+s^{2}+4s+2)}+{(2s^{3}+2s^{2}+2s)}$ =$\ G'(x)={4s^{3}+3s^{2}+6s+2}$

10) Una caja con tapa se fabricará con una hoja rectangular de cartón, que mide 5 por 8 pies. Esto se realiza cortando las regiones sombreadas de la figura y luego doblando por las líneas discontinuas. ¿ Cuales son las dimensiones x, y, z que maximizan el volumen?

$\ z={5-2x}$ $\ 2y={8-2x}$ =$\ y={\frac {(2)(4-x)}{2}}$ =$\ y={4-x}$

$\ Area={(L)(L)(L)}$ =$\ A={(x)(5-2x)(4-x)}$ =$\ A={2x^{3}-13x^{2}+20x}$ =$\ {\frac {dA}{dx}}={(3)(2x^{3-1})-(2)(13x^{2-1})+(20)}$ =$\ {\frac {dA}{dx}}={6x^{2}-26x+20}$ =$\ {\frac {dA}{dx}}={(6)6x^{2}-(26)6x+(6)20}$ =$\ {\frac {dA}{dx}}={36x^{2}-(26)6x+120}$ =$\ {\frac {dA}{dx}}={(6x-20)(6x-6)}$ =$\ x={\frac {20}{6}}={\frac {10}{3}}$ =$\ x={\frac {6}{6}}={1}$

=$\ x={1}$ =$\ y={4-(1)}={3}$ =$\ z={5-2(1)}={3}$

Proximamente mas..................

200.71.53.4</math>