User:Corredorclau

DERIVADAS
1)Derivar:$$F(x)=x^4-7x^2+6$$
 * $$F'(x)=4x^3-14x$$

--Corredorclau 14:08, 12 Jul 2004 (UTC)

2)Derivar:$$F(x)=5x^2-2x-7$$
 * $$F'(x)=10x-2$$

--Corredorclau 14:15, 12 Jul 2004 (UTC)

3)Derivar:$$F(x)=(2x^-5)^2$$
 * $$F'(x)=2(2x^2-5)(4x)$$
 * $$F'(x)=(4x^2-10)(4x)$$
 * $$F'(x)=16x^3-40

by: corredorclau$$

Los siguientes ejercicios fueron tomados del libro de Howard E. Taylor.

4)Derivar:$$ f(x)=\frac {3}{x}+\frac {7}{x^2}-x^{-3}$$
 * $$ f'(x)=\frac {x-3}{x^2}+\frac {x^2-(2x)(7)}{x^4}+3x^{-4}$$
 * $$ f'(x)=\frac {-3}{x^2}-\frac {14x}{(x^4}+3x^{-4}$$
 * $$ f'(x)=\frac {-3}{x^2}-\frac {14}{(x^3}+\frac{3}{x^4}$$

Corredorclau 18:18, 13 Jul 2004 (UTC)

5)Derivar:$$F(x)= 3x^{\frac {1}{3}}+7senx$$
 * $$F'(x)= \frac {1}{3}+3x^{\frac {1}{3}-1}+7cosx$$
 * $$F'(x)= 3x^{\frac {-2}{3}}+7cosx$$

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6)Derivar:$$F(x)= x\sqrt{5+x^2}$$
 * $$F'(x)= \sqrt{5+x^2}+x\cdot \frac {1}{2\sqrt{5+x^2}}\cdot 2x$$
 * $$F'(x)= \sqrt{5+x^2}+\frac {2x^2}{2\sqrt{5+x^2}}$$
 * $$F'(x)= \frac {\sqrt{5+x^2}+x^2}{\sqrt{5+x^2}}$$
 * $$F'(x)= \frac {5+x^2+x^2}{\sqrt{5+x^2}}$$
 * $$F'(x)= \frac {5+2x^2}{\sqrt{5+x^2}}$$

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7)Derivar:$$\ 3x^2+4y^2=12$$
 * $$\ 6x+8yy'=0$$
 * $$\ 8yy'=6x$$
 * $$\ y'=\frac {6x}{8y}$$
 * $$\ y'=\frac {-3x}{4y}$$

Corredorclau 21:44, 13 Jul 2004 (UTC)

8)Derivar:$$\ F(x)=(x^2-4)(x^4+5)$$
 * $$\ F'(x)=(2x)(x^4+5)+(x^5-4)(4x^3)$$
 * $$\ F'(x)=2x^5+10x+4x^5-16x^3$$
 * $$\ F'(x)=6x^5-16x^3+10x$$

Corredorclau 12:48, 13 Jul 2004 (UTC)

9)Derivar:$$\ F(x)=2x^2(2x-4)$$
 * $$\ F'(x)=4x(2x-4)+2x^2(2)$$
 * $$\ F'(x)=8x^2-16x+4x^2$$
 * $$\ F'(x)=12x2-16x$$

Corredorclau 12:48, 13 Jul 2004 (UTC)

10)Derivar:$$\ y(x)=3x^3-4x^2+5x+16$$
 * $$\ y'(x)=9x^2-8x+5+0$$
 * $$\ y'(x)=9x^2-8x+5$$

JORGE MARIO MEDINA MARTIN 5:10, 13 Jul 2004

11):$$ f(x)=\frac {senx}{x}$$
 * $$ f'(x)=\frac {cosx(x)-senx}{x^2}$$
 * $$ f'(x)=\frac {cosx}{x}-\frac {senx}{x^2}$$

Corredorclau

12)Derivar:$$ f(x)=\frac {2x^3+4x}{3senx}$$
 * $$ f'(x)=\frac {(6X^2-4(3senx))-(2x^3-4x)3cosx}{(3senx)^2}$$
 * $$ f'(x)=\frac {18x^2senx-12senx-6x^3cosx+12xcosx}{9sen^2x}$$

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13)Derivar:$$ f(x)=\frac {1}{(x^2-2x)^2}$$
 * $$ f'(x)=\frac {2(x^2-2x)(2x-2)}{(x^2-2x)^4}$$
 * $$ f'(x)=\frac {4x-4}{(x^2-2x)^3}$$

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14)Derivar:$$ f(x)=\frac {1-senx}{1+cosx}$$
 * $$ f'(x)=\frac {-cosx(1+cosx)-(1-senx)-senx}{1+cosx}$$
 * $$ f'(x)=\frac {-cosx-cos^2x+senx-sen^2x}{1+cos^2x}$$
 * $$ f'(x)=\frac {-cosx-1+sen^2x+senx-sen^2x}{(1+cosx)^2}$$
 * $$ f'(x)=\frac {-cosx-1+senx}{(1+cosx)^2}$$

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15)Derivar:$$ f(x)=\frac {cos^22x}{x^2}$$
 * $$ f'(x)=\frac {2(cos2x)sen2x(2)(x^2)-2x(cos^22x)}{x^4}$$
 * $$ f'(x)=\frac {4senx-2x}{x^2}$$

Corredorclau

16)Derivar:$$ f(x)=\frac {cosx}{7}$$
 * $$ f'(x)=\frac {-7senx}{49}$$
 * $$ f'(x)=\frac {-senx}{7}$$

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17)Derivar:$$ f(x)=\frac {(x-1)^2}{x^2}$$
 * $$ f'(x)=\frac {2(x-1)(x^2)-(x-1)^22x}{x^4}$$
 * $$ f'(x)=\frac {(2x-2)(x^2)-(x^2-2x+1)2x}{x^4}$$
 * $$ f'(x)=\frac {2x^3-^2x^2-2x^3+4x^2-2x}{x^4}$$
 * $$ f'(x)=\frac {2x^2-2x}{x^4}$$
 * $$ f'(x)=\frac {2x(x-1)}{x^4}$$
 * $$ f'(x)=\frac {2(x-1)}{x^3}$$

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18)Derivar:$$ f(x)=x+(\frac {1}{x})^2$$
 * $$ f(x)=2(x+\frac {1}{x})(1-\frac {-1}{x^2})$$
 * $$ f(x)=(2x+\frac {2}{x})(1-(\frac {1}{x^2})$$
 * $$ f(x)=(\frac {2x^2+2}{x})(\frac {x^2-1}{x^2})$$
 * $$ f(x)=\frac {2x^2-2x^2+2x^2-2}{x^3}$$
 * $$ f(x)=\frac {2x^4-2}{x^3}$$

Corredorclau

19)derivar:$$f(x)=cos^5(2x^3+4)$$
 * $$f'(x)=5cos^4(2x^3+4)(6x^2)$$
 * $$f'(x)=30x^2cos^4(2x^3+4)$$

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20)derivar:$$f(x)=sen7x$$
 * $$f'(x)=cos7x(7)$$
 * $$f'(x)=7cox7x$$

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21)derivar y calcular f'(1):$$f(t)=sen(t^2+3t+1)$$
 * $$f'(t)=cos(t^2+3t+1)(2t+3)$$
 * $$f'(1)=cos25$$

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22) derivar y hallar g'(1):$$g(t)=(t^2+9)^3(t^2-2)^4$$
 * $$g'(t)=3(t^2+9)(2t)(t^2-2)^4+(t^2+9)^3(4)(t^2-2)^3(2t)$$
 * $$g'(t)=(3t^2+27)^2(2t^3-4t)^4+(t^2+9)^3(4)(t^2-2)^3(2t)$$
 * $$g'(1)=-113600$$

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23)encontrar la recta tangente de
 * $$y=(x^2+1)^3(x^4+1)^2 en los puntos (1,32)$$
 * $$y'=3(x^2+1)(2x)(x^4+1)^2+(x^2+1)^32(x^4+1)(4x^3)$$
 * $$evaluo a x=1 en y'$$
 * $$224$$
 * $$la ecuacion de la recta es y=224x-192$$

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24) derivar:$$f(x)=5x^2+4x+\frac{7}{x^5}+8cosx$$
 * $$f'(x)=10x+4+\frac{35x^4}{x^10}-8senx$$
 * $$f'(x)=10x+4+\frac{35}{x^6}-8senx$$

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SEGUNDA DERIVADA
25)halle la segunda derivada de:$$f(x)=xsenx$$
 * $$f'(x)=senx+xcosx$$
 * $$f''(x)=cosx+cosx-xsenx$$
 * $$f''(x)=2cosx-xsenx$$

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26)halle la segunda derivada de$$f(x)=(1+x)^15$$
 * $$f'(x)=15(1+x)^14$$
 * $$f''(x)=210(1+x)^13$$

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27):$$F(x)=(7+x)^5$$
 * $$f'(x)=5(7+x)^4$$
 * $$f''(x)=2087+x)^3$$

Corredorclau

28):$$f(x)=(3-2x)^5$$
 * $$f'(x)=5(3-2x)^4(-2)$$
 * $$f'(x)=-10(3-2x)^4$$
 * $$f''(x)=-40(3-2x)^3(-2)$$
 * $$f''(x)=80(3-2x)^3$$

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29):$$f/(x)=(4+2x^2)^7$$
 * $$f'(x)=7(4+2x^2)^6(4x)$$
 * $$f'(x)=28x(4+2x^2)^6$$
 * $$f''(x)=28(4+2x^2)^6+28x(4+2x^2)^5(4x)$$
 * $$f''(x)=12(4+2x^2)^6+112x^2(4+2x^2)^5$$

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DERIVACION IMPLICITA
30):$$3X^2+4Y^2=12$$
 * $$6X+8YY'=0$$
 * $$8YY'=-6X$$
 * $$Y'=\frac{6x}{8y}$$

Corredorclau

31):$$x^3+y^3-3x^2+3y^2=0$$
 * $$3x^2+3y^2y'-6x+6x+6yy'=0$$
 * $$3y^2y'+6yy'=-3x^2+6x$$
 * $$y'(3y^2+6y)=-3x^2+6x$$
 * $$y'=-\frac{3x^2+6y}{3y^2+6y}$$

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32):$$senx+cosy=0$$
 * $$cosx-senyy'=0$$
 * $$-senyy'=-cosx$$
 * $$y'=\frac{cosx}{seny}$$

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33):$$4x^3+7xy^2=2y^3$$
 * $$12x^2+7y^2+7x2yy'=6y^2y'$$
 * $$12x^2+7y^2=6y^2y'-7x2yy'$$
 * $$12x^2+7y^2=y'(6y^2-7x2y)$$
 * $$\frac{12x^2+7y^2}{6y^2-7x2y}$$

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DERIVADAS por regla de la cadena
Es utilizada para derivar funciones compuestas,primero se deriva la compuesta y se multiplica por la interna.

34):$$f(t)=(\frac{3t-2}{t+5})^3$$
 * $$f'(t)=3(\frac{3t-2}{t+5})^2(\frac{3(t+5)-(3t-2)}{(t+5)^2})$$
 * $$f'(t)=3(\frac{3t-2}{t+5})^2(\frac{17}{(t+5)^2)}$$

Corredorclau

35):$$f(t)=\frac{83t-2)^3}{t+5}$$
 * $$f'(t)=\frac{3(3t-2)^23(t+5)-(3t-2)^3}{(t+5)^2}$$
 * $$f'(t)=\frac{6t+30(3t-2)^2-(3t-2)^3}{(t+5)^2}$$

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36):$$f(x)=sen^3x$$
 * $$f'(x)=3cosx^2(cos)$$

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37):$$f(t)=(senttan(t^2+1))$$
 * $$f´(t)=(costtan(t^2+1)+sentsec^2(t^2+1)(2t))$$
 * $$f'(t)=(costtan(t^2+1)+sentsec^2(2t^3+2t))$$

Corredorclau

38):$$H(x)=\sqrt{x^2-1}$$
 * $$H'(x)=\frac{1}{2\sqrt{x^2-1}}(2x)$$
 * $$H'(x)=\frac{x}{\sqrt{x^2-1}}$$

Corredorclau

39):$$H(x)=sen^2x+senx^2$$
 * $$2senxcosx+cosx^2(2x)$$

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MÁXIMOS Y MINIMOS
1) Al lado de un gran muro de una granja. se quiere cercar un terreno que tenga forma rectangular. Se dispone solamente de 100 metros de malla de alambre para construir la cerca para que esta encierre la mayor parte de área posible.


 * $$A=xy$$
 * $$2x+y=100$$
 * $$y=100-2x$$
 * $$A=x(100+2x)$$
 * $$A=100x-2x^2$$
 * $$A'=100-4x$$
 * $$A'=0$$
 * $$100-4x=0$$
 * $$-4x=-100$$
 * $$x=\frac {100}{4}$$
 * $$x=25$$


 * $$y=100-2x$$
 * $$y=100-50$$
 * $$y=25$$


 * $$A=xy$$
 * $$A=25(50)$$
 * $$A=1250 metros$$

Corredorclau

''2) cual es la maxima area que puede tener un triangulo rectangulo cuya hipotenusa tenga 5cm de largo.

Por pitagoras
 * $$Z^2=x^2+y^2$$
 * $$z=5$$
 * $$x^2+y^2=25$$
 * $$y=\sqrt{25-x^2}$$

area del triangulo
 * $$A=\frac{xy}{2}$$
 * $$A=x(\frac{\sqrt{25-x^2}}{2})$$
 * $$\frac{1}{2}\sqrt{25-x^2}+(\frac({x}{2})(\frac{-2x}{2(\sqrt{25-x^2})}=0$$
 * $$\frac{25-x^2-x^2}{2\sqrt{25-x^2}}=0$$
 * $$25-2x^2=0$$
 * $$x=\sqrt{\frac{25}{2}}$$
 * $$y=\sqrt{25-\frac{25}{2}}$$
 * $$y=\sqrt{\frac{25}{2}}$$
 * $$A=\frac{\sqrt{\frac{25}{2}}}{\sqrt{\frac{25}{2}}}{2}$$
 * $$A=\frac{25}{4}cm^2$$

Corredorclau''