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

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

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?

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

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

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

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200.71.53.4