User talk:"Lina Rincon"

Limites
Estos ejercicios fueron tomados de un taller

Evalue los siguientes limites

1. $$\lim_{x\to 1} \frac {x^2+2x-3} {x^2-5x+4}$$ =$$\lim_{x\to 1} \frac {(x+3)(x-1)} {(x-1)(x-4)}$$ =$$\lim_{x\to 1} \frac {x+3} {x-4}$$ =$$\frac {1+3} {1-4}$$ =$$\frac {4} {-3}$$ 2. $$\lim_{x\to 1} \frac {x^4-x^5} {1-x}$$ =$$\lim_{x\to 1} \frac {x^4(1-x)} {1-x}$$ =$$\lim_{x\to 1} \frac {x^4}{1}$$ =$$\ 1^4 $$    =$$\ 1 $$

Derivadas
Estos ejercicios fueren tomados del calculo de Purcell, octava edicion.

1. En el problema esta dibujada una recta tangente a una curva, evalue su pendiente.

$$\lim_{h\to 0} \frac {f(c+h) - f(c)} {h}$$
 * Respuesta:
 * $$\lim_{h\to 0}\frac {(5+h)- (5)} {h}$$
 * $$\lim_{h\to 0}\frac {25+10+h^2-25} {h}$$
 * $$\lim_{h\to 0}\frac {10h +h^2} {h}$$
 * $$\lim_{h\to 0}\frac {h(10 +h)} {h}$$
 * $$\ {10+0} $$
 * $$\ 10 $$
 * "Lina Rincon"

2. Encuentre las pendientes de las recta tangentes a la curva $$\ y = x^2-1 $$ en los puntos en donde x = -2, -1, 0, 1, 2.


 * $$\lim_{h\to 0} \frac {f(c+h) - f(c)} {h}$$
 * Respuesta:
 * $$\lim_{h\to 0} \frac {((c+h)^2-1) - (c^2-1)} {h}$$
 * $$\lim_{h\to 0}\frac {c^2+2ch+h^2-1-c^2+1} {h}$$
 * $$\lim_{h\to 0}\frac {2ch+h^2} {h}$$
 * $$\lim_{h\to 0}\frac {h(2c+h)} {h}$$
 * $$\ {2c+0} $$
 * $$\ 2c $$
 * (2.-2=-4), (2.-1=-2),  (2.0=0),  (2.1=2),  (2.2=4)
 * "Lina Rincon"

3. En los problemas siguientes utilice la definicíón $$f`(x)=lim_{h\to 0}\frac{f(c+h)-f(c)}{h}$$
 * a. f(1) si f(x)=x^2/


 * Respuesta:


 * $$\lim_{h\to 0}\frac{f(1+h)-f(1)}{h}$$
 * $$\lim_{h\to 0} \frac {(1+h)^2-(1)^2} {h}$$
 * $$\lim_{h\to 0} \frac {1+2h+h^2-1} {h}$$
 * $$\lim_{h\to 0} \frac {2h+h^2} {h}$$
 * $$\lim_{h\to 0} \frac {h(2+h)} {h}$$
 * $$\ 2+0 $$
 * $$\ 2 $$


 * b. f(2) si f(t)=(2t)^2/


 * Respuesta:


 * $$\lim_{h\to 0} \frac {f(c+h) - f(c)} {h}$$
 * $$\lim_{h\to 0} \frac {f(2+h) - f(2)} {h}$$
 * $$\lim_{h\to 0}\frac {2(2+h)^2-2(2)^2}{h}$$
 * $$\lim_{h\to 0}\frac{(2(4+4h+h^2)-2(4)}{h}$$
 * $$\lim_{h\to 0} \frac {8+8h+2h^2-8} {h}$$
 * $$\lim_{h\to 0} \frac {8h+2h^2} {h}$$
 * $$\lim_{h\to 0} \frac {h(8+2h)} {h}$$
 * $$\ {8+0}$$
 * $$\ 8$$


 * c. f(3) si f(t)=t^2-t/


 * Respuesta:


 * $$\lim_{h\to 0} \frac {f(c+h) - f(c)} {h}$$
 * $$\lim_{h\to 0} \frac {f(3+h) - f(3)} {h}$$
 * $$\lim_{h\to 0}\frac{((3+h)^2-(3+h))-(3^2-3)}{h}$$
 * $$\lim_{h\to 0} \frac {(9+6h+h^2-3-h)-(9-3)} {h}$$
 * $$\lim_{h\to 0} \frac {9+6h+h^2-3-h-6)} {h}$$
 * $$\lim_{h\to 0} \frac {5h+h^2)} {h}$$
 * $$\lim_{h\to 0} \frac {h(5+h)} {h}$$
 * $$\ {5+0}$$
 * $$\ 5$$

4.En los siguientes problemas utilice las reglas para encontrar derivadas.

a. $$\ Y = {2x^2} $$
 * Respuesta:
 * $$\ Dx Y= {2.2x^{2-1}} $$
 * $$\ Dx Y= {4x} $$

"Lina Rincon"

b. $$\ Y = {3x^3} $$
 * Respuesta:
 * $$\ Dx Y = {3.3x^{2-1}} $$
 * $$\ Dx Y = {9x^2} $$

"Lina Rincon"

c. $$\ Y = {pi.x} $$
 * $$\ pi DxY = {x} $$
 * $$\ pi DxY = {1} $$
 * $$\ DxY = {pi.1} $$
 * $$\ DxY = {pi} $$

"Lina Rincon"

d. $$\ Y = {pi.x^3} $$
 * $$\ piDxY = {3.1x^{3-1}} $$
 * $$\ piDxY = {3x^2}$$
 * $$\ DxY = {3pix^2}$$

"Lina Rincon"

e. $$\ DxY = \frac {500x^4} {x^10}$$
 * $$\ DxY = {2.(-2)x^{-2-1}} $$
 * $$\ DxY = {-4x^{-3}} $$

"Lina Rincon"

f. $$\ Y = {-3x^{-4}} $$
 * $$\ DxY = {-4.(-3)x^{-4-1}} $$
 * $$\ DxY = {-12x^{-5}} $$

"Lina Rincon"

g. $$\ Y = \frac {pi} {x}$$
 * $$\ DxY = \frac {(0.pi)-(1.pi)} {x^2}$$
 * $$\ DxY = \frac {0-pi} {x^2}$$
 * $$\ DxY = \frac {-pi} {x^2}$$

"Lina Rincon"

h. $$\ Y = \frac {100} {x^5}$$
 * $$\ DxY = \frac {(0.x^5)-(100.5x^4)} {{(x^5)}^2}$$
 * $$\ DxY = \frac {0-500x^4} {x^10}$$
 * $$\ DxY = \frac {500x^4} {x^10}$$
 * $$\ DxY = \frac {500} {x^6}$$

"Lina Rincon"

i. $$\ Y = {x^2+2x} $$
 * $$\ DxY = {Dx(x^2)+Dx(2x)} $$
 * $$\ DxY = {Dx(2x)+Dx(2)} $$
 * $$\ DxY = {2x+2} $$

"Lina Rincon"

j. $$\ Y = {3x^4+x^3} $$
 * $$\ DxY = {Dx(3x^4)+Dx(x^3)} $$
 * $$\ DxY = {Dx(12x^3)+Dx(3x^2)} $$
 * $$\ DxY = {12x^3+3x^2} $$

"Lina Rincon"

i. $$\ Y = {x^4+x^3+x^2+x+1} $$
 * $$\ DxY = {Dx(x^4)+Dx(x^3)+Dx(x^2)+Dx(x)+Dx(1)} $$
 * $$\ DxY = {Dx(4x^3)+Dx(3x^2)+Dx(2x)+Dx(1)+Dx(0)} $$
 * $$\ DxY = {4x^3+3x^2+2x+1} $$

"Lina Rincon"