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)$$\ y = 9-x^3;(2,5)$$
 * $$\ m(x)= lim_{h\to 0}\frac {9-(x+h)^2-(9-x^2)}{h}$$
 * $$\ m(x)= lim_{h\to 0}\frac {9-(x^2+2xh+h^2)-9+x^2}{h}$$
 * $$\ m(x)= lim_{h\to 0}\frac {9-x^2-2xh-h^2-9+x^2}{h}$$
 * $$\ m(x)= lim_{h\to 0}\frac {-2xh-h^2}{h}$$
 * $$\ m(x)= lim_{h\to 0}\frac {(h)(-2x-h)}{h}$$
 * $$\ m(x)= lim_{h\to 0} -2x-h$$
 * $$\ m(x)= -2(2)-(0)$$
 * $$\ m(x)= -4$$


 * $$\ y-5=-4(x-2)$$
 * $$\ y-5=-4x+8$$
 * $$\ y=-4x+8+5$$
 * $$\ y=-4x+13$$

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


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


 * $$\ y-0=-4(x-(-2))$$
 * $$\ y=-4x-8$$

3)$$\ y=x^3+3;(1,4)$$
 * $$\ m(x)= lim_{h\to 0}\frac {(x+h)^3+3-(x^3+3)}{h}$$
 * $$\ m(x)= lim_{h\to 0}\frac {x^3+3x^2h+3xh^2+h^3+3-x^3-3)}{h}$$
 * $$\ m(x)= lim_{h\to 0}\frac {3x^2h+3xh^2+h^3)}{h}$$
 * $$\ m(x)= lim_{h\to 0}\frac {(h)(3x^2+3xh+h^2)}{h}$$
 * $$\ m(x)= lim_{h\to 0} 3x^2+3xh+h^2$$
 * $$\ m(x)= 3(1)^2+3(1)(0)+(0)^2$$
 * $$\ m(x)= 3$$


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

II)Determine la pendiente de la recta tangente en el punto $$(x_1,f(x_1))$$ y dónde es horizontal

1)$$\ f(x)= 3x^2-12x+8$$
 * $$\ f(x_1)= 3x_1^2-12x_1+8$$
 * $$\ f(x_1+h)= 3(x_1+h)^2-12(x_1+h)+8$$


 * $$\ m(x)= lim_{h\to 0}\frac {(3(x_1+h)^2-12(x_1+h)+8) - (3x_1^2-12x_1+8)}{h}$$
 * $$\ m(x)= lim_{h\to 0}\frac {(3(x_1^2+2x_1h+h^2)-12x_1-12h)+8) - (3x_1^2-12x_1+8)}{h}$$
 * $$\ m(x)= lim_{h\to 0}\frac {3x_1^2+6x_1h+3h^2-12x_1-12h+8 - 3x_1^2+12x_1-8)}{h}$$
 * $$\ m(x)= lim_{h\to 0}\frac {6x_1h+3h^2-12h}{h}$$
 * $$\ m(x)= lim_{h\to 0}\frac {(h)(6x_1+3h-12)}{h}$$
 * $$\ m(x)= lim_{h\to 0}6x_1+3h-12$$
 * $$\ m(x)= 6x_1+3(0)-12$$
 * $$\ m(x)= 6x_1-12$$

Horizontal


 * $$\ 6x_1-12=0$$
 * $$\ 6x_1=12$$
 * $$\ x_1=\frac {12}{6}$$
 * $$\ x_1=2$$


 * $$\ y=3(2^2)-12(2)+8$$
 * $$\ y=-4$$

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


 * $$\ f(x_1)= x_1^3+6x_1^2-9x_1-2$$
 * $$\ f(x_1+h)= (x_1+h)^3+6(x_1+h)^2-9(x_1+h)-2$$


 * $$\ m(x)= lim_{h\to 0}\frac {((x_1+h)^3+6(x_1+h)^2-9(x_1+h)-2) - (x_1^3+6x_1^2-9x_1-2)}{h}$$
 * $$\ m(x)= lim_{h\to 0}\frac {3x_1^2h+3x_1h^2+h^3-12x_1h-6h^2-9h}{h}$$
 * $$\ m(x)= lim_{h\to 0}\frac {(h)(3x_1^2+3x_1h+h^2-12x_1-6h-9)}{h}$$
 * $$\ m(x)= lim_{h\to 0}3x_1^2+3x_1h+h^2-12x_1-6h-9$$
 * $$\ m(x)= 3x_1^2+3x_1(0)+(0)^2-12x_1-6(0)-9$$
 * $$\ m(x)= 3x_1^2-12x_1-9$$

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) $$\lim_{x\to 0}\frac {(x-sen(x))^2}{x^2}$$
 * =$$\lim_{x\to 0}\frac {x^2-2xsen(x)+sen^2x}{x^2}$$
 * =$$\lim_{x\to 0}\frac {x^2}{x^2} - \lim_{x\to 0}\frac {2xsen^2x}{x^2}+ \lim_{x\to 0}\frac {sen^2x}{x^2}$$
 * =$$\ 1 - \lim_{x\to 0}\frac {2x}{x}\cdot\frac {senx}{x}+\frac {senx}{x}\cdot\lim_{x\to 0}\frac {senx}{x}$$
 * =$$\ 1 - (2)(1) + (1)(1)$$
 * =$$\ 1 - 2 + 1$$
 * =$$\ 0$$
 * Marcelaburaglia

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

DERIVACIÓN IMPLÍCITA
1)$$\ Sen\left( \frac {x}{y}\right)+Cos\left( \frac {y}{x}\right)=0$$
 * $$\ Cos\left( \frac {x}{y}\right)\cdot \left( \frac {y-x\frac {dy}{dx}}{y^2}\right)-Sen \left( \frac {y}{x}\right)\cdot \left( \frac {x\frac {dy}{dx}-y}{x^2}\right)=0$$
 * $$\ \frac {Cos\left( \frac {x}{y}\right)}{y^2}\cdot \left( y-x\frac {dy}{dx}\right) - \frac {Sen\left( \frac {y}{x}\right)}{x^2}\cdot \left(x\frac {dy}{dx}-y\right)=0$$
 * $$\ \frac {x^2[Cos\left( \frac {x}{y}\right)\cdot(y-x \frac {dy}{dx})]- y^2[Sen\left( \frac {y}{x}\right)\cdot(x\frac {dy}{dx}-y)]}{x^2y^2}=0$$
 * $$\ x^2[yCos\left( \frac {x}{y}\right)-x\frac {dy}{dx}Cos\left( \frac {x}{y}\right)] - y^2[x\frac {dy}{dx}Sen\left( \frac {y}{x}\right)\cdot ySen\frac {y}{x}] = 0\cdot x^2y^2$$
 * $$\ x^2yCos\left(\frac {x}{y}\right)-x^3\frac {dy}{dx}Cos\left(\frac {x}{y}\right) - xy^2\frac {dy}{dx}Sen\left(\frac {y}{x}\right)+y^3Sen\left(\frac {y}{x}\right) = 0$$
 * $$\ -x^3\frac {dy}{dx} Cos\left(\frac {x}{y}\right) - xy^2\frac {dy}{dx}Sen\left(\frac {y}{x}\right) = -x^2yCos\left(\frac {x}{y}\right)-y^3Sen\left(\frac {y}{x}\right)$$
 * $$\ \frac {dy}{dx}\left(-x^3Cos\left(\frac {x}{y}\right) - xy^2Sen\left(\frac {y}{x}\right)\right) = -x^2yCos\left(\frac {x}{y}\right)-y^3Sen\left(\frac {y}{x}\right)$$
 * $$\ \frac {dy}{dx}= \frac {-x^2yCos\left(\frac {x}{y}\right)-y^3Sen\left(\frac {y}{x}\right)}{-x^3Cos\left(\frac {x}{y}\right) - xy^2Sen\left(\frac {y}{x}\right)} $$
 * $$\ \frac {dy}{dx}= \frac {y\left(-x^2Cos\left(\frac {x}{y}\right)-y^2Sen\left(\frac {y}{x}\right)\right)}{x \left(-x^2Cos\left(\frac {x}{y}\right) - y^2Sen\left(\frac {y}{x}\right)\right)} $$
 * $$\ \frac {dy}{dx}= \frac {y}{x}$$


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

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

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