# User:Victorpollo

1) ${\displaystyle lim_{x\to 0}{\frac {3xtanx}{senx}}}$

=${\displaystyle lim_{x\to 0}{\frac {3xsenx/cosx}{senx}}}$
=${\displaystyle lim_{x\to 0}{\frac {3xsenx}{senxcosx}}}$
=${\displaystyle {\frac {0}{1}}}$
=${\displaystyle \ 0}$

2) ${\displaystyle lim_{x\to 0}{\frac {sen3x}{2x}}}$

=${\displaystyle 3/3lim_{x\to 0}{\frac {sen3x}{2x}}}$
=$\displaystyle 3*lim_{x\to 0}\1/3\frac {sen3x}{2x}$
=${\displaystyle 2/2*3lim_{x\to 0}{\frac {senx}{2x}}}$
=${\displaystyle 3/2lim_{x\to 0}{\frac {senx}{x}}}$
=${\displaystyle {\frac {2}{3}}}$

3)Derive: ${\displaystyle \ y=3x^{2}-1/4x+3}$

${\displaystyle y'={\frac {6x(4x+3)-(3x^{2}-1)4}{(4x+3)^{2}}}}$
${\displaystyle y'={\frac {12x^{2}+18x+4}{(4x+3)^{2}}}}$

4)Derive:${\displaystyle \ y=1/2x(4-x^{5})+{\frac {3x-1}{x}}}$

${\displaystyle y'=1/2x(-5x^{4})+(4-x^{5})1/2+{\frac {3*x-(3x-1)*1}{x^{2}}}}$
=${\displaystyle -5/2x^{5}+2-1/2x^{5}+{\frac {3x-3x+1}{x^{2}}}}$
=${\displaystyle -3x^{5}+2+{\frac {1}{x^{2}}}}$
${\displaystyle y'={\frac {-3x^{7}+2x^{2}+1}{x^{2}}}}$

5)Derive:${\displaystyle \ y={\frac {2x^{2}-4x+3}{2-3x}}}$

${\displaystyle y^{=}{\frac {(4x-4)(2-3x)-(2x^{2}-4x+3)(-3)}{(2-3x)^{2}}}}$
=${\displaystyle {\frac {(-12x^{2}+20x-8)-(-6x^{2}+12x-9)}{(2-3x)^{2}}}}$
${\displaystyle y'(x)={\frac {-6x^{2}+8x+1}{(2-3x)^{2}}}}$

6)Derive:${\displaystyle \ y=(3x-2x^{2})(5+4x)}$

${\displaystyle y'=(3x-2x^{2})(4)+(5+4x)(3-4x)}$
${\displaystyle y'=12x-8x^{2})+(15-8x-16X^{2})}$
${\displaystyle y'=-24x^{2}+4x+15}$

hallar primera y segunda derivada 7)${\displaystyle y=x^{3}+3x^{2}+6x}$

${\displaystyle y'=3x^{2}+6x+6}$
${\displaystyle y''=6x+6}$

8)${\displaystyle y=senx^{3}}$

${\displaystyle y'=(cosx^{3})(3x^{2})}$
${\displaystyle y''=(-senx^{3}*3x^{2})(3x^{2})+(cosx^{3})(6x)}$
${\displaystyle y''=9x^{4}senx^{3}+6xcosx^{3}}$

9)${\displaystyle y={\frac {3x}{1-x}}}$

${\displaystyle y'={\frac {(3)(1-x)-(3x)(-1)}{(1-x)^{2}}}}$
${\displaystyle y'={\frac {3-3x+3x}{(1-x)^{2}}}}$
${\displaystyle y'={\frac {3}{(1-x)^{2}}}}$
${\displaystyle y''={\frac {(0)(1-x)^{2}-(3)(2(x-1)(-1))}{(1-x)^{4}}}}$
${\displaystyle y''={\frac {6(1-x)}{(1-x)^{4}}}}$
${\displaystyle y''={\frac {6}{(1-x^{3})}}}$

10)${\displaystyle y=(3x+5)^{3}}$

${\displaystyle Y'=3(3x+5)^{2}(3)}$
${\displaystyle y'=9(3x+5)^{2}}$
${\displaystyle y''=18(3x+5)(3)}$
${\displaystyle y''=54(3x+5)}$
${\displaystyle y''=162x+270}$

Derivar implicitamente 11)${\displaystyle 2xy=(x^{2}+y^{2})^{3}/2}$

${\displaystyle Dx(2xy)=Dx(x^{2}+y^{2})^{3}/2}$
${\displaystyle (2x)(y')+2y=3/2(x^{2}+y^{2})^{1}/2(2x+yy')}$
${\displaystyle 2xy'+2y=3x(x^{2}+y^{2})^{1}/2+3(x^{2}+y^{2})^{1}/2yy'}$
${\displaystyle 2xy'-3(x^{2}+y^{2})^{1}/2yy'=3(x^{2}+y^{2})^{1}/2-2y}$
${\displaystyle y'(2x-3y/x^{2}+y^{2})^{1}/2))=3x(x^{2}+y^{2})^{1}/2-2y}$
${\displaystyle y'={\frac {3x(x^{2}+y^{2})^{1}/2-2y}{2x-3y/x^{2}+y^{2})^{1}/2}}}$

12)${\displaystyle y^{3}+y^{2}-5y-x^{2}=-4}$

${\displaystyle 3y^{2}y'+2yy'-5y'-2x=0}$
${\displaystyle y'(3y^{2}+2y-5)=2x}$
${\displaystyle y'={\frac {2x}{3y^{2}+2y-5}}}$

13)${\displaystyle xy+xy^{2}=6}$

${\displaystyle xy'+y*1+x*2yy'+y^{2}*1=0}$
${\displaystyle xy'+y+2xyy'+y^{2}=0}$
${\displaystyle y'(x+2xy)=-y-y^{2}}$
${\displaystyle y'={\frac {-}{y+y^{2}}}{x+2xy}}$

14)${\displaystyle y^{2}=x^{4}+x^{2}y^{2}}$

${\displaystyle 2yy'=4x^{3}+x^{2}*2yy'+y^{2}*2x}$
${\displaystyle y'(2y-2x^{2}y)=4x^{3}+2xy^{2}}$
${\displaystyle y'={\frac {4x^{3}+2xy^{2}}{2y-2x^{2}y}}}$

15)${\displaystyle xseny+cos2y=cosy}$

${\displaystyle (x)(cosyy')+(seny)+(-sen2y)*2y'=(-senyy')}$
${\displaystyle xcosyy'-2(sen2y)y'+senyy'=-seny}$
${\displaystyle y'(xcosy-2seny+seny)=-seny}$
${\displaystyle y'={\frac {-seny}{xcosy-2seny+seny}}}$

Problemas de razones de cambio relacionadas

16)De un tubo sale arena a razon de 16 pies^3/seg, si la arena al caer siempre forma un cono cuya altura siempre es 1/4 del tamaño de la base que tan rapido aumenta h cuando el monton tiene una h de 4 pies.

ANS: se sabe que h= 4 pies, y nos preguntan dh/dt es decir la razon de cambio de la altura con respecto al tiempo y tambien se sabe que 16pies^3/seg es dv/dt es decir la razon de cambio del volumen con respecto al tiempo.

${\displaystyle V=1/3pir^{2}h}$
${\displaystyle V=1/3pi(2h)^{2}h}$
${\displaystyle V=1/3pi(4h^{2})h}$
${\displaystyle {\frac {dv}{dt}}={\frac {4pi}{3}}*3h^{2}{\frac {dh}{dt}}}$
${\displaystyle {\frac {dv}{dt}}=4pi*h^{2}{\frac {dh}{dt}}}$
${\displaystyle {\frac {dh}{dt}}={\frac {dv}{dt}}*{\frac {1}{4pih^{2}}}}$
${\displaystyle {\frac {dh}{dt}}=16pies^{3}/seg*{\frac {1}{64pi}}}$
${\displaystyle {\frac {dh}{dt}}=0.0795pies^{3}/seg}$