Victorpollo

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

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

=$lim_{x\to 0}{\frac {3xsenx}{senxcosx}}$

=${\frac {0}{1}}$

=$\ 0$

Victorpollo

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

=$3/3lim_{x\to 0}{\frac {sen3x}{2x}}$

=Failed to parse (syntax error): {\displaystyle 3*lim_{x\to 0}\1/3\frac {sen3x}{2x}}

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

=$3/2lim_{x\to 0}{\frac {senx}{x}}$

=${\frac {2}{3}}$

Victorpollo

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

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

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

Victorpollo

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

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

=$-5/2x^{5}+2-1/2x^{5}+{\frac {3x-3x+1}{x^{2}}}$

=$-3x^{5}+2+{\frac {1}{x^{2}}}$

$y'={\frac {-3x^{7}+2x^{2}+1}{x^{2}}}$

Victorpollo

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

$y^{=}{\frac {(4x-4)(2-3x)-(2x^{2}-4x+3)(-3)}{(2-3x)^{2}}}$

=${\frac {(-12x^{2}+20x-8)-(-6x^{2}+12x-9)}{(2-3x)^{2}}}$

$y'(x)={\frac {-6x^{2}+8x+1}{(2-3x)^{2}}}$

Victorpollo

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

$y'=(3x-2x^{2})(4)+(5+4x)(3-4x)$

$y'=12x-8x^{2})+(15-8x-16X^{2})$

$y'=-24x^{2}+4x+15$

Victorpollo

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

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

$y''=6x+6$

Victorpollo

8)$y=senx^{3}$

$y'=(cosx^{3})(3x^{2})$

$y''=(-senx^{3}*3x^{2})(3x^{2})+(cosx^{3})(6x)$

$y''=9x^{4}senx^{3}+6xcosx^{3}$

Victorpollo

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

$y'={\frac {(3)(1-x)-(3x)(-1)}{(1-x)^{2}}}$

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

$y'={\frac {3}{(1-x)^{2}}}$

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

$y''={\frac {6(1-x)}{(1-x)^{4}}}$

$y''={\frac {6}{(1-x^{3})}}$

Victorpollo

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

$Y'=3(3x+5)^{2}(3)$

$y'=9(3x+5)^{2}$

$y''=18(3x+5)(3)$

$y''=54(3x+5)$

$y''=162x+270$

Victorpollo

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

$Dx(2xy)=Dx(x^{2}+y^{2})^{3}/2$

$(2x)(y')+2y=3/2(x^{2}+y^{2})^{1}/2(2x+yy')$

$2xy'+2y=3x(x^{2}+y^{2})^{1}/2+3(x^{2}+y^{2})^{1}/2yy'$

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

$y'(2x-3y/x^{2}+y^{2})^{1}/2))=3x(x^{2}+y^{2})^{1}/2-2y$

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

Victorpollo

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

$3y^{2}y'+2yy'-5y'-2x=0$

$y'(3y^{2}+2y-5)=2x$

$y'={\frac {2x}{3y^{2}+2y-5}}$

Victorpollo

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

$xy'+y*1+x*2yy'+y^{2}*1=0$

$xy'+y+2xyy'+y^{2}=0$

$y'(x+2xy)=-y-y^{2}$

$y'={\frac {-}{y+y^{2}}}{x+2xy}$

Victorpollo

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

$2yy'=4x^{3}+x^{2}*2yy'+y^{2}*2x$

$y'(2y-2x^{2}y)=4x^{3}+2xy^{2}$

$y'={\frac {4x^{3}+2xy^{2}}{2y-2x^{2}y}}$

Victorpollo

15)$xseny+cos2y=cosy$

$(x)(cosyy')+(seny)+(-sen2y)*2y'=(-senyy')$

$xcosyy'-2(sen2y)y'+senyy'=-seny$

$y'(xcosy-2seny+seny)=-seny$

$y'={\frac {-seny}{xcosy-2seny+seny}}$

Victorpollo

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.

$V=1/3pir^{2}h$

$V=1/3pi(2h)^{2}h$

$V=1/3pi(4h^{2})h$

${\frac {dv}{dt}}={\frac {4pi}{3}}*3h^{2}{\frac {dh}{dt}}$

${\frac {dv}{dt}}=4pi*h^{2}{\frac {dh}{dt}}$

${\frac {dh}{dt}}={\frac {dv}{dt}}*{\frac {1}{4pih^{2}}}$

${\frac {dh}{dt}}=16pies^{3}/seg*{\frac {1}{64pi}}$

${\frac {dh}{dt}}=0.0795pies^{3}/seg$

Victorpollo