# User:Corredorclau

1)Derivar:${\displaystyle F(x)=x^{4}-7x^{2}+6}$

${\displaystyle F'(x)=4x^{3}-14x}$

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

2)Derivar:${\displaystyle F(x)=5x^{2}-2x-7}$

${\displaystyle F'(x)=10x-2}$

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

3)Derivar:${\displaystyle F(x)=(2x^{-}5)^{2}}$

${\displaystyle F'(x)=2(2x^{2}-5)(4x)}$
${\displaystyle F'(x)=(4x^{2}-10)(4x)}$
${\displaystyle F'(x)=16x^{3}-40~~~by:corredorclau}$

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

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

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

Corredorclau 18:18, 13 Jul 2004 (UTC)

5)Derivar:${\displaystyle F(x)=3x^{\frac {1}{3}}+7senx}$

${\displaystyle F'(x)={\frac {1}{3}}+3x^{{\frac {1}{3}}-1}+7cosx}$
${\displaystyle F'(x)=3x^{\frac {-2}{3}}+7cosx}$

6)Derivar:${\displaystyle F(x)=x{\sqrt {5+x^{2}}}}$

${\displaystyle F'(x)={\sqrt {5+x^{2}}}+x\cdot {\frac {1}{2{\sqrt {5+x^{2}}}}}\cdot 2x}$
${\displaystyle F'(x)={\sqrt {5+x^{2}}}+{\frac {2x^{2}}{2{\sqrt {5+x^{2}}}}}}$
${\displaystyle F'(x)={\frac {{\sqrt {5+x^{2}}}+x^{2}}{\sqrt {5+x^{2}}}}}$
${\displaystyle F'(x)={\frac {5+x^{2}+x^{2}}{\sqrt {5+x^{2}}}}}$
${\displaystyle F'(x)={\frac {5+2x^{2}}{\sqrt {5+x^{2}}}}}$

7)Derivar:${\displaystyle \ 3x^{2}+4y^{2}=12}$

${\displaystyle \ 6x+8yy'=0}$
${\displaystyle \ 8yy'=6x}$
${\displaystyle \ y'={\frac {6x}{8y}}}$
${\displaystyle \ y'={\frac {-3x}{4y}}}$

Corredorclau 21:44, 13 Jul 2004 (UTC)

8)Derivar:${\displaystyle \ F(x)=(x^{2}-4)(x^{4}+5)}$

${\displaystyle \ F'(x)=(2x)(x^{4}+5)+(x^{5}-4)(4x^{3})}$
${\displaystyle \ F'(x)=2x^{5}+10x+4x^{5}-16x^{3}}$
${\displaystyle \ F'(x)=6x^{5}-16x^{3}+10x}$

Corredorclau 12:48, 13 Jul 2004 (UTC)

9)Derivar:${\displaystyle \ F(x)=2x^{2}(2x-4)}$

${\displaystyle \ F'(x)=4x(2x-4)+2x^{2}(2)}$
${\displaystyle \ F'(x)=8x^{2}-16x+4x^{2}}$
${\displaystyle \ F'(x)=12x2-16x}$

Corredorclau 12:48, 13 Jul 2004 (UTC)

10)Derivar:${\displaystyle \ y(x)=3x^{3}-4x^{2}+5x+16}$

${\displaystyle \ y'(x)=9x^{2}-8x+5+0}$
${\displaystyle \ y'(x)=9x^{2}-8x+5}$

JORGE MARIO MEDINA MARTIN 5:10, 13 Jul 2004

11):${\displaystyle f(x)={\frac {senx}{x}}}$

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

12)Derivar:${\displaystyle f(x)={\frac {2x^{3}+4x}{3senx}}}$

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

13)Derivar:${\displaystyle f(x)={\frac {1}{(x^{2}-2x)^{2}}}}$

${\displaystyle f'(x)={\frac {2(x^{2}-2x)(2x-2)}{(x^{2}-2x)^{4}}}}$
${\displaystyle f'(x)={\frac {4x-4}{(x^{2}-2x)^{3}}}}$

14)Derivar:${\displaystyle f(x)={\frac {1-senx}{1+cosx}}}$

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

15)Derivar:${\displaystyle f(x)={\frac {cos^{2}2x}{x^{2}}}}$

${\displaystyle f'(x)={\frac {2(cos2x)sen2x(2)(x^{2})-2x(cos^{2}2x)}{x^{4}}}}$
${\displaystyle f'(x)={\frac {4senx-2x}{x^{2}}}}$

16)Derivar:${\displaystyle f(x)={\frac {cosx}{7}}}$

${\displaystyle f'(x)={\frac {-7senx}{49}}}$
${\displaystyle f'(x)={\frac {-senx}{7}}}$

17)Derivar:${\displaystyle f(x)={\frac {(x-1)^{2}}{x^{2}}}}$

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

18)Derivar:${\displaystyle f(x)=x+({\frac {1}{x}})^{2}}$

${\displaystyle f(x)=2(x+{\frac {1}{x}})(1-{\frac {-1}{x^{2}}})}$
${\displaystyle f(x)=(2x+{\frac {2}{x}})(1-({\frac {1}{x^{2}}})}$
${\displaystyle f(x)=({\frac {2x^{2}+2}{x}})({\frac {x^{2}-1}{x^{2}}})}$
${\displaystyle f(x)={\frac {2x^{2}-2x^{2}+2x^{2}-2}{x^{3}}}}$
${\displaystyle f(x)={\frac {2x^{4}-2}{x^{3}}}}$

19)derivar:${\displaystyle f(x)=cos^{5}(2x^{3}+4)}$

${\displaystyle f'(x)=5cos^{4}(2x^{3}+4)(6x^{2})}$
${\displaystyle f'(x)=30x^{2}cos^{4}(2x^{3}+4)}$

20)derivar:${\displaystyle f(x)=sen7x}$

${\displaystyle f'(x)=cos7x(7)}$
${\displaystyle f'(x)=7cox7x}$

21)derivar y calcular f'(1):${\displaystyle f(t)=sen(t^{2}+3t+1)}$

${\displaystyle f'(t)=cos(t^{2}+3t+1)(2t+3)}$
${\displaystyle f'(1)=cos25}$

22) derivar y hallar g'(1):${\displaystyle g(t)=(t^{2}+9)^{3}(t^{2}-2)^{4}}$

${\displaystyle g'(t)=3(t^{2}+9)(2t)(t^{2}-2)^{4}+(t^{2}+9)^{3}(4)(t^{2}-2)^{3}(2t)}$
${\displaystyle g'(t)=(3t^{2}+27)^{2}(2t^{3}-4t)^{4}+(t^{2}+9)^{3}(4)(t^{2}-2)^{3}(2t)}$
${\displaystyle g'(1)=-113600}$

23)encontrar la recta tangente de

${\displaystyle y=(x^{2}+1)^{3}(x^{4}+1)^{2}enlospuntos(1,32)}$
${\displaystyle y'=3(x^{2}+1)(2x)(x^{4}+1)^{2}+(x^{2}+1)^{3}2(x^{4}+1)(4x^{3})}$
${\displaystyle evaluoax=1eny'}$
${\displaystyle 224}$
${\displaystyle laecuaciondelarectaesy=224x-192}$

24) derivar:${\displaystyle f(x)=5x^{2}+4x+{\frac {7}{x^{5}}}+8cosx}$

${\displaystyle f'(x)=10x+4+{\frac {35x^{4}}{x^{1}0}}-8senx}$
${\displaystyle f'(x)=10x+4+{\frac {35}{x^{6}}}-8senx}$

25)halle la segunda derivada de:${\displaystyle f(x)=xsenx}$

${\displaystyle f'(x)=senx+xcosx}$
${\displaystyle f''(x)=cosx+cosx-xsenx}$
${\displaystyle f''(x)=2cosx-xsenx}$

26)halle la segunda derivada de${\displaystyle f(x)=(1+x)^{1}5}$

${\displaystyle f'(x)=15(1+x)^{1}4}$
${\displaystyle f''(x)=210(1+x)^{1}3}$

27):${\displaystyle F(x)=(7+x)^{5}}$

${\displaystyle f'(x)=5(7+x)^{4}}$
${\displaystyle f''(x)=2087+x)^{3}}$

28):${\displaystyle f(x)=(3-2x)^{5}}$

${\displaystyle f'(x)=5(3-2x)^{4}(-2)}$
${\displaystyle f'(x)=-10(3-2x)^{4}}$
${\displaystyle f''(x)=-40(3-2x)^{3}(-2)}$
${\displaystyle f''(x)=80(3-2x)^{3}}$

29):${\displaystyle f/(x)=(4+2x^{2})^{7}}$

${\displaystyle f'(x)=7(4+2x^{2})^{6}(4x)}$
${\displaystyle f'(x)=28x(4+2x^{2})^{6}}$
${\displaystyle f''(x)=28(4+2x^{2})^{6}+28x(4+2x^{2})^{5}(4x)}$
${\displaystyle f''(x)=12(4+2x^{2})^{6}+112x^{2}(4+2x^{2})^{5}}$

## DERIVACION IMPLICITA

30):${\displaystyle 3X^{2}+4Y^{2}=12}$

${\displaystyle 6X+8YY'=0}$
${\displaystyle 8YY'=-6X}$
${\displaystyle Y'={\frac {6x}{8y}}}$

31):${\displaystyle x^{3}+y^{3}-3x^{2}+3y^{2}=0}$

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

32):${\displaystyle senx+cosy=0}$

${\displaystyle cosx-senyy'=0}$
${\displaystyle -senyy'=-cosx}$
${\displaystyle y'={\frac {cosx}{seny}}}$

33):${\displaystyle 4x^{3}+7xy^{2}=2y^{3}}$

${\displaystyle 12x^{2}+7y^{2}+7x2yy'=6y^{2}y'}$
${\displaystyle 12x^{2}+7y^{2}=6y^{2}y'-7x2yy'}$
${\displaystyle 12x^{2}+7y^{2}=y'(6y^{2}-7x2y)}$
${\displaystyle {\frac {12x^{2}+7y^{2}}{6y^{2}-7x2y}}}$

## DERIVADAS por regla de la cadena

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

34):${\displaystyle f(t)=({\frac {3t-2}{t+5}})^{3}}$

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

35):${\displaystyle f(t)={\frac {83t-2)^{3}}{t+5}}}$

${\displaystyle f'(t)={\frac {3(3t-2)^{2}3(t+5)-(3t-2)^{3}}{(t+5)^{2}}}}$
${\displaystyle f'(t)={\frac {6t+30(3t-2)^{2}-(3t-2)^{3}}{(t+5)^{2}}}}$

36):${\displaystyle f(x)=sen^{3}x}$

${\displaystyle f'(x)=3cosx^{2}(cos)}$

37):${\displaystyle f(t)=(senttan(t^{2}+1))}$

$\displaystyle f´(t)=(costtan(t^2+1)+sentsec^2(t^2+1)(2t))$
${\displaystyle f'(t)=(costtan(t^{2}+1)+sentsec^{2}(2t^{3}+2t))}$

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

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

39):${\displaystyle H(x)=sen^{2}x+senx^{2}}$

${\displaystyle 2senxcosx+cosx^{2}(2x)}$

## 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.

${\displaystyle A=xy}$
${\displaystyle 2x+y=100}$
${\displaystyle y=100-2x}$
${\displaystyle A=x(100+2x)}$
${\displaystyle A=100x-2x^{2}}$
${\displaystyle A'=100-4x}$
${\displaystyle A'=0}$
${\displaystyle 100-4x=0}$
${\displaystyle -4x=-100}$
${\displaystyle x={\frac {100}{4}}}$
${\displaystyle x=25}$
${\displaystyle y=100-2x}$
${\displaystyle y=100-50}$
${\displaystyle y=25}$
${\displaystyle A=xy}$
${\displaystyle A=25(50)}$
${\displaystyle A=1250metros}$

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

Por pitagoras

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

area del triangulo

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