# User:Angelicapulido

DERIVAR LAS SIGUIENTES FUNCIONES

1) ${\displaystyle \ f(x)=10x^{2}-9x-4}$

${\displaystyle \ f'(x)=20x+9}$

2) ${\displaystyle \ f(x)=6x^{3}-5x^{2}+x-9}$

${\displaystyle \ f'(x)=18x^{2}-10x+1}$

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

${\displaystyle \ f'(x)=(x^{3}-7)(4x)+(2x^{2}+3)(3x^{2})}$
${\displaystyle \ f'(x)=4x^{4}-28x+6x^{4}+9x^{2}}$
${\displaystyle \ f'(x)=10x^{4}+9x^{2}-28x}$

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

${\displaystyle \ f'(x)=(x^{2})(12x^{3}-7)+(3x^{4}-7x+2)(2x)}$
${\displaystyle \ f'(x)=12x^{5}-7x^{2}+6x^{5}-14x^{2}+4x}$
${\displaystyle \ f'(x)=18x^{5}-21x^{2}+4x}$

5)${\displaystyle \ f(x)={\frac {4x-5}{3x+2}}}$

${\displaystyle \ f'(x)={\frac {(3x+2)(4)-(4x-5)(3)}{(3x+2)^{2}}}}$
${\displaystyle \ f'(x)={\frac {12x+8-12x+15}{(3x+2)^{2}}}}$
${\displaystyle \ f'(x)={\frac {23}{(3x+2)^{2}}}}$

6)${\displaystyle \ f(x)={\frac {8-x+3x^{2}}{2-9x}}}$

${\displaystyle \ f'(x)={\frac {(2-9x)(6x-1)-(8-x+3x^{2})(-9)}{(2-9x)^{2}}}}$
${\displaystyle \ f'(x)={\frac {12x-2-54x^{2}+9x+72-9x+27x^{2}}{(2-9x)^{2}}}}$
${\displaystyle \ f'(x)={\frac {12x-27x^{2}+70}{(2-9x)^{2}}}}$

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

${\displaystyle \ f'(x)={\frac {(1+x+x^{2}+x^{3})(0)-(1)(1+2x+3x^{2})}{(1+x+x^{2}+x^{3})^{2}}}}$
${\displaystyle \ f'(x)={\frac {0-1+2x+3x^{2}}{(1+x+x^{2}+x^{3})^{2}}}}$
${\displaystyle \ f'(x)={\frac {-3x^{2}-2x-1}{(1+x+x^{2}+x^{3})^{2}}}}$

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

${\displaystyle \ f'(x)={\frac {(x^{2})(3)-(3x-1)(2x)}{(x^{2})^{2}}}}$
${\displaystyle \ f'(x)={\frac {3x^{2}-6x^{2}-2x}{(x^{4}}}}$
${\displaystyle \ f'(x)={\frac {-3x^{2}-2x}{(x^{4})}}}$

9)${\displaystyle \ x^{2}+y^{2}=1}$

${\displaystyle \ 2x+2y{\frac {dy}{dx}}=0}$
${\displaystyle \ 2y{\frac {dy}{dx}}=-2x}$
${\displaystyle \ {\frac {dy}{dx}}={\frac {-2x}{2y}}}$
${\displaystyle \ {\frac {dy}{dx}}={\frac {-x}{y}}}$

10)${\displaystyle \ y^{2}={\frac {x-1}{x+1}}}$

${\displaystyle \ 2y{\frac {dy}{dx}}={\frac {(x+1)-(x-1)}{(x+1)^{2}}}}$
${\displaystyle \ 2y{\frac {dy}{dx}}={\frac {x+1-x+1}{(x+1)^{2}}}}$
${\displaystyle \ {\frac {dy}{dx}}={\frac {2}{2y(x+1)^{2}}}}$
${\displaystyle \ {\frac {dy}{dx}}={\frac {1}{y(x+1)^{2}}}}$

11)${\displaystyle \ x^{2}+xy=2}$

${\displaystyle \ 2x+x{\frac {dy}{dx}}+y=0}$
${\displaystyle \ x{\frac {dy}{dx}}=-2x-y}$
${\displaystyle \ {\frac {dy}{dx}}={\frac {-2x+y}{x}}}$

12)${\displaystyle \ x^{2}y+xy^{2}=6}$

${\displaystyle \ x^{2}{\frac {dy}{dx}}+2xy+2xy{\frac {dy}{dx}}+y^{2}=0}$
${\displaystyle \ x^{2}{\frac {dy}{dx}}+2xy{\frac {dy}{dx}}=-2xy-y^{2}}$
${\displaystyle \ {\frac {dy}{dx}}(x^{2}+2xy)=-2xy-y^{2}}$
${\displaystyle \ {\frac {dy}{dx}}={\frac {-2xy-y^{2}}{x^{2}+2xy}}}$

13)${\displaystyle \ {\frac {1}{y}}+{\frac {1}{x}}=1}$

${\displaystyle \ y^{-}1+x^{-}1=0}$
${\displaystyle \ -y^{-}2{\frac {dy}{dx}}-x^{-}2=0}$
${\displaystyle \ -y^{-}2{\frac {dy}{dx}}=x^{-}2}$
${\displaystyle \ {\frac {dy}{dx}}={\frac {x^{-}2}{-y^{-}2}}}$

14)${\displaystyle \ y^{2}=x^{2}(x^{2}+1)}$

${\displaystyle \ 2y{\frac {dy}{dx}}=2x(x^{2}+1)+(2x)(x^{2})}$
${\displaystyle \ 2y{\frac {dy}{dx}}=2x(x^{2}+1+x^{2})}$
${\displaystyle \ {\frac {dy}{dx}}={\frac {2x(x^{2}+1+x^{2})}{2y}}}$
${\displaystyle \ {\frac {dy}{dx}}={\frac {x(2x^{2}+1)}{y}}}$

15)${\displaystyle \ x^{2}y^{2}=x^{2}+y^{2}}$

${\displaystyle \ 2yx^{2}{\frac {dy}{dx}}+2xy^{2}=2x+2y{\frac {dy}{dx}}}$
${\displaystyle \ 2yx^{2}{\frac {dy}{dx}}-2y{\frac {dy}{dx}}=2x-2xy^{2}}$
${\displaystyle \ 2y{\frac {dy}{dx}}(x^{2}-1)=2x-2xy^{2}}$
${\displaystyle \ {\frac {dy}{dx}}={\frac {2x-2xy^{2}}{2y(x^{2}-1)}}}$
${\displaystyle \ {\frac {dy}{dx}}={\frac {2x(1-y^{2})}{2y(x^{2}-1)}}}$
${\displaystyle \ {\frac {dy}{dx}}={\frac {x(1-y^{2})}{y(x^{2}-1)}}}$

HALLAR EL LIMITE DE LAS SIGUIENTES FUNCIONES

dada :${\displaystyle \ f(x)=x^{2}-3x}$ hallar:${\displaystyle \lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}}$

${\displaystyle \lim _{h\to 0}{\frac {(x+h)^{2}-3(x+h)-(x^{2}-3x)}{h}}}$
${\displaystyle \lim _{h\to 0}{\frac {x^{2}+2xh+h^{2}-3x-3h-x^{2}+3x}{h}}}$
${\displaystyle \lim _{h\to 0}{\frac {2xh+h^{2}-3h}{h}}}$
${\displaystyle \lim _{h\to 0}2x+h-3=2x-3}$