User:Marcelaburaglia

RECTA TANGENTE EN UN PUNTO

I. Obtener la ecuación de la recta tangente en el punto dado. (Ejercicios tomados del libro: 7 ed. de Leithold)

1)$\ y=9-x^{3};(2,5)$

$\ m(x)=lim_{h\to 0}{\frac {9-(x+h)^{2}-(9-x^{2})}{h}}$

$\ m(x)=lim_{h\to 0}{\frac {9-(x^{2}+2xh+h^{2})-9+x^{2}}{h}}$

$\ m(x)=lim_{h\to 0}{\frac {9-x^{2}-2xh-h^{2}-9+x^{2}}{h}}$

$\ m(x)=lim_{h\to 0}{\frac {-2xh-h^{2}}{h}}$

$\ m(x)=lim_{h\to 0}{\frac {(h)(-2x-h)}{h}}$

$\ m(x)=lim_{h\to 0}-2x-h$

$\ m(x)=-2(2)-(0)$

$\ m(x)=-4$

$\ y-5=-4(x-2)$

$\ y-5=-4x+8$

$\ y=-4x+8+5$

$\ y=-4x+13$

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2)$\ y=2x^{2}+4x;(-2,0)$

$\ m(x)=lim_{h\to 0}{\frac {2(x+h)^{2}+4(x+h)-(2x^{2}+4x)}{h}}$

$\ m(x)=lim_{h\to 0}{\frac {2(x^{2}+2xh+h^{2})+4x+4h-2x^{2}-4x}{h}}$

$\ m(x)=lim_{h\to 0}{\frac {2x^{2}+4xh+2h^{2}+4x+4h-2x^{2}-4x}{h}}$

$\ m(x)=lim_{h\to 0}{\frac {4xh+2h^{2}+4h}{h}}$

$\ m(x)=lim_{h\to 0}{\frac {(h)(4x+2h+4)}{h}}$

$\ m(x)=lim_{h\to 0}4x+2h+4$

$\ m(x)=4(-2)+2(0)+4$

$\ m(x)=-4$

$\ y-0=-4(x-(-2))$

$\ y=-4x-8$

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3)$\ y=x^{3}+3;(1,4)$

$\ m(x)=lim_{h\to 0}{\frac {(x+h)^{3}+3-(x^{3}+3)}{h}}$

$\ m(x)=lim_{h\to 0}{\frac {x^{3}+3x^{2}h+3xh^{2}+h^{3}+3-x^{3}-3)}{h}}$

$\ m(x)=lim_{h\to 0}{\frac {3x^{2}h+3xh^{2}+h^{3})}{h}}$

$\ m(x)=lim_{h\to 0}{\frac {(h)(3x^{2}+3xh+h^{2})}{h}}$

$\ m(x)=lim_{h\to 0}3x^{2}+3xh+h^{2}$

$\ m(x)=3(1)^{2}+3(1)(0)+(0)^{2}$

$\ m(x)=3$

$\ y-4=3(x-1)$

$\ y-4=3x-3$

$\ y=3x-3+4$

$\ y=3x+1$

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II)Determine la pendiente de la recta tangente en el punto $(x_{1},f(x_{1}))$ y dónde es horizontal

1)$\ f(x)=3x^{2}-12x+8$

$\ f(x_{1})=3x_{1}^{2}-12x_{1}+8$

$\ f(x_{1}+h)=3(x_{1}+h)^{2}-12(x_{1}+h)+8$

$\ m(x)=lim_{h\to 0}{\frac {(3(x_{1}+h)^{2}-12(x_{1}+h)+8)-(3x_{1}^{2}-12x_{1}+8)}{h}}$

$\ m(x)=lim_{h\to 0}{\frac {(3(x_{1}^{2}+2x_{1}h+h^{2})-12x_{1}-12h)+8)-(3x_{1}^{2}-12x_{1}+8)}{h}}$

$\ m(x)=lim_{h\to 0}{\frac {3x_{1}^{2}+6x_{1}h+3h^{2}-12x_{1}-12h+8-3x_{1}^{2}+12x_{1}-8)}{h}}$

$\ m(x)=lim_{h\to 0}{\frac {6x_{1}h+3h^{2}-12h}{h}}$

$\ m(x)=lim_{h\to 0}{\frac {(h)(6x_{1}+3h-12)}{h}}$

$\ m(x)=lim_{h\to 0}6x_{1}+3h-12$

$\ m(x)=6x_{1}+3(0)-12$

$\ m(x)=6x_{1}-12$

Horizontal

$\ 6x_{1}-12=0$

$\ 6x_{1}=12$

$\ x_{1}={\frac {12}{6}}$

$\ x_{1}=2$

$\ y=3(2^{2})-12(2)+8$

$\ y=-4$

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2)$\ f(x)=x^{3}+6x^{2}-9x-2$

$\ f(x_{1})=x_{1}^{3}+6x_{1}^{2}-9x_{1}-2$

$\ f(x_{1}+h)=(x_{1}+h)^{3}+6(x_{1}+h)^{2}-9(x_{1}+h)-2$

$\ m(x)=lim_{h\to 0}{\frac {((x_{1}+h)^{3}+6(x_{1}+h)^{2}-9(x_{1}+h)-2)-(x_{1}^{3}+6x_{1}^{2}-9x_{1}-2)}{h}}$

$\ m(x)=lim_{h\to 0}{\frac {3x_{1}^{2}h+3x_{1}h^{2}+h^{3}-12x_{1}h-6h^{2}-9h}{h}}$

$\ m(x)=lim_{h\to 0}{\frac {(h)(3x_{1}^{2}+3x_{1}h+h^{2}-12x_{1}-6h-9)}{h}}$

$\ m(x)=lim_{h\to 0}3x_{1}^{2}+3x_{1}h+h^{2}-12x_{1}-6h-9$

$\ m(x)=3x_{1}^{2}+3x_{1}(0)+(0)^{2}-12x_{1}-6(0)-9$

$\ m(x)=3x_{1}^{2}-12x_{1}-9$

Horizontal

$\ 3x_{1}^{2}-12x_{1}-9=0$

$\ (x_{1}-3)(3x+3)=0$

$\ x_{1}=3$

$\ x_{1}=1$

$\ y_{1}=3(1)^{2}-12(1)-9$

$\ y_{1}=0$

$\ y_{3}=3(3)^{2}-12(3)-9$

$\ y_{3}=-56$

DERIVADAS TRIGONOMÉTRICAS

1) $\lim _{x\to 0}{\frac {(x-sen(x))^{2}}{x^{2}}}$

=$\lim _{x\to 0}{\frac {x^{2}-2xsen(x)+sen^{2}x}{x^{2}}}$

=$\lim _{x\to 0}{\frac {x^{2}}{x^{2}}}-\lim _{x\to 0}{\frac {2xsen^{2}x}{x^{2}}}+\lim _{x\to 0}{\frac {sen^{2}x}{x^{2}}}$

=$\ 1-\lim _{x\to 0}{\frac {2x}{x}}\cdot {\frac {senx}{x}}+{\frac {senx}{x}}\cdot \lim _{x\to 0}{\frac {senx}{x}}$

=$\ 1-(2)(1)+(1)(1)$

=$\ 1-2+1$

=$\ 0$

Marcelaburaglia

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MÁXIMOS Y MÍNIMOS

Estime los números críticos de cada función y luego encuentre los valores máximos y mínimos

1)$\ f(x)=x^{3}+7x^{2}-5x$

$\ f'(x)=x^{3}+7x^{2}-5x$

$\ f'(x)=3x^{2}+14x-5$

$\ f'(x)=(x+5)(3x-1)$

$\ x=-5$

$\ x={\frac {1}{3}}$

$\ f_{(-5)}=(-5)^{3}+7(-5)^{2}-5(-5)$

$\ f_{(-5)}=-125+175+25$

$\ f_{(-5)}=75$ Máximo

$\ f_{({\frac {1}{3}})}=({\frac {1}{3}})^{3}+7({\frac {1}{3}})^{2}-5({\frac {1}{3}})$

$\ f_{({\frac {1}{3}})}={\frac {1}{27}}+{\frac {7}{9}}-{\frac {5}{3}}$

$\ f_{({\frac {1}{3}})}={\frac {-23}{27}}$ Mínimo

Marcelaburaglia

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DERIVACIÓN IMPLÍCITA

1)$\ Sen\left({\frac {x}{y}}\right)+Cos\left({\frac {y}{x}}\right)=0$

$\ Cos\left({\frac {x}{y}}\right)\cdot \left({\frac {y-x{\frac {dy}{dx}}}{y^{2}}}\right)-Sen\left({\frac {y}{x}}\right)\cdot \left({\frac {x{\frac {dy}{dx}}-y}{x^{2}}}\right)=0$

$\ {\frac {Cos\left({\frac {x}{y}}\right)}{y^{2}}}\cdot \left(y-x{\frac {dy}{dx}}\right)-{\frac {Sen\left({\frac {y}{x}}\right)}{x^{2}}}\cdot \left(x{\frac {dy}{dx}}-y\right)=0$

$\ {\frac {x^{2}[Cos\left({\frac {x}{y}}\right)\cdot (y-x{\frac {dy}{dx}})]-y^{2}[Sen\left({\frac {y}{x}}\right)\cdot (x{\frac {dy}{dx}}-y)]}{x^{2}y^{2}}}=0$

$\ x^{2}[yCos\left({\frac {x}{y}}\right)-x{\frac {dy}{dx}}Cos\left({\frac {x}{y}}\right)]-y^{2}[x{\frac {dy}{dx}}Sen\left({\frac {y}{x}}\right)\cdot ySen{\frac {y}{x}}]=0\cdot x^{2}y^{2}$

$\ x^{2}yCos\left({\frac {x}{y}}\right)-x^{3}{\frac {dy}{dx}}Cos\left({\frac {x}{y}}\right)-xy^{2}{\frac {dy}{dx}}Sen\left({\frac {y}{x}}\right)+y^{3}Sen\left({\frac {y}{x}}\right)=0$

$\ -x^{3}{\frac {dy}{dx}}Cos\left({\frac {x}{y}}\right)-xy^{2}{\frac {dy}{dx}}Sen\left({\frac {y}{x}}\right)=-x^{2}yCos\left({\frac {x}{y}}\right)-y^{3}Sen\left({\frac {y}{x}}\right)$

$\ {\frac {dy}{dx}}\left(-x^{3}Cos\left({\frac {x}{y}}\right)-xy^{2}Sen\left({\frac {y}{x}}\right)\right)=-x^{2}yCos\left({\frac {x}{y}}\right)-y^{3}Sen\left({\frac {y}{x}}\right)$

$\ {\frac {dy}{dx}}={\frac {-x^{2}yCos\left({\frac {x}{y}}\right)-y^{3}Sen\left({\frac {y}{x}}\right)}{-x^{3}Cos\left({\frac {x}{y}}\right)-xy^{2}Sen\left({\frac {y}{x}}\right)}}$

$\ {\frac {dy}{dx}}={\frac {y\left(-x^{2}Cos\left({\frac {x}{y}}\right)-y^{2}Sen\left({\frac {y}{x}}\right)\right)}{x\left(-x^{2}Cos\left({\frac {x}{y}}\right)-y^{2}Sen\left({\frac {y}{x}}\right)\right)}}$

$\ {\frac {dy}{dx}}={\frac {y}{x}}$

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• $\ Sen(u)=Sen(Sen^{-1}(x))$

$\ Sen(y)=x$

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3)$\ y=Sen^{-1}(x)$

$\ Sen(y)=x$

$\ Cos(y){\frac {dy}{dx}}=1$

$\ {\frac {dy}{dx}}={\frac {1}{Cos(y)}}$

$\ {\frac {dy}{dx}}={\frac {1}{\sqrt {1-Sen^{2}(x)}}}$

$\ {\frac {dy}{dx}}={\frac {1}{\sqrt {1-x^{2}}}}$

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4)$\ y=Cos^{-1}(x)$

$\ Cos(y)=x$

$\ -Sen(y){\frac {dy}{dx}}=1$

$\ {\frac {dy}{dx}}={\frac {-1}{Sen(y)}}$

$\ {\frac {dy}{dx}}={\frac {-1}{\sqrt {1-Cos^{2}x}}}$

$\ {\frac {dy}{dx}}={\frac {-1}{\sqrt {1-x^{2}}}}$