# User:IvanH

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

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

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

4) $\displaystyle \ f(r)= /frac {4πr^3}{3}$ =$\displaystyle \ f'(r)= /frac {(3)4πr^{3-1}}{3}$ =$\displaystyle \ f'(r)= {4πr^2}$

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

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

7) ${\displaystyle \ g(x)={x^{2}+{\frac {1}{x^{2}}}}}$ =$\displaystyle \ g'(x)= {(2)(x^{2-1}) + \frac {(0)(x^2)-(1)(2)(x^{2-1})}{(x^2)^2}$ =${\displaystyle \ g'(x)={2x}+{\frac {0-2x}{x^{4}}}}$ =${\displaystyle \ g'(x)={2x}-{\frac {2}{x^{3}}}}$

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

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

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?

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

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

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

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

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