x 2 x 2 f x ax b 2 x 2 2 x 6 x2 alim x 2 x 2 lim x 2 ax b 2 2 2 x b b 2 a 4 b

X 2 x 2 f x ax b 2 x 2 2 x 6 x2 alim x 2 x 2 lim x 2

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x2x≤2f(x)ax+b2<x<22x6x≥2a)limx →2x2=limx →2ax+b(2)2=−2x+bb2a=4b)limx→2ax+b=limx →22x62a+b=462a+b=−2
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b2a=4b=1b+2a=−2a=322b=212.2xa x3f(x)ax+2b3≤ x≤3b5b x>3a)limx →32xa=limx →3ax+2b2(3)a=a(3)+2b6a=−3a+2b2b2a=−6ab=3b)limx→3ax+2b=limx→3b5x3a+2b=b153a+b=−153a+b=−15a=−3ab=3b=−64a=−12
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Calculo diferencial1.2.f(x)=x6+x125xf '(x)=6x5+12x125f '(x)=6x5+12x53.u(x)=1x+1x2+1x3u'(x)=x1+x2+x3u'(x)=−1x22x33x4u'(x)=1x22x33x44.u(x)=x2+x3x5x=x2x+x3xx5xu'(r)=x32+x52x2dudx=32x321+52x5212x21dudx=32x12+52x322x5.f(t)=t5+t7t95f(t)=t5+t72t95dfdt=5t51+72t72195t951dfdt=5t4+72t5295t456.g(x)=x12+x13+x14
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g(x)=12x121+13x131+14x141g(x)=12x12+13x23+14x34=12x+133x2+144x37.f(x)=i=1ni x1if(x)=x+2x12+3x13+4x14+...+n x1ndfdx=1+212x12+3¿13x23+414x34+...+n1nx1nndfdx=1+x12+x23+x34+...+x1nndfdx=i=1nx1ii8.¿r2+1¿¿r¿7r3¿u(r)=¿¿r¿7r77r71(r3)'(r2+1)r3(r2+1)'(r2+1)2dudr=¿¿r¿6r¿r77r63r2(r2+1)r32r(r2+1)2¿dudr=¿¿r¿r3r4+3r22r4(r2+1)2¿r¿6¿dudr=7¿¿r¿r3r2+r4(r2+1)2¿r¿6¿dudr=7¿9.g(t)=t5+t47t4+t16
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dudr=(5t4+47t471)(t4+t16)(t5+t47)(4t3+16t56)(t4+t16)2dudr=(5t4+47t37)(t4+t16)(t4+t16)2(t5+t47)(4t3+16t56)(t4+t16)2dudr=5t4+47t37t4+t16(t5+t47)(4t3+16t56)(t4+t16)210.h(s)=4s54+7s66s6+75dh(s)ds=4(54)s541+7(6s5) (6s7+7)7s6(67s6)(6s7+7)2dh(s)ds=5s14+252s12+294s5294s12(6s7+7)2dh(s)ds=5s14+294s5425s12(6s7+7)211.u(x)=|x2x|dudx=|x2x|x2x(2x1)12.v(x)=|x51|+|x|x2+5xdv(x)dx=(|x51|+|x|x515x4+|x|x)(x2+5x)(|x51|+|x|)(2x+5)(x2+5x)2dv(x)d x=|x51|+|x|x515x4+|x|xx2+5x(|x51|+|x|)(2x+5)(x2+5x)2
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4.2derivada aplicando la definición 1.y=x+x2f(x)=x+x2y'=limh→0f(x+h)f(x)hf(x+h)=x+h+(x+h)2y'=limh→0x+h+(x+h)2xx2hy'=limh→0h+x2+2xh+h2x2h=limh→0h(1+2x+h)hy'=1+2x2.y=2x+3x3f(x)=2x+3x3y'=limh→0f(x+h)f(x)hf(x+h)=2(x+h)+3(x+h)3y'=limh→02x+2h+3(x3+3x2h+3x h2+h3)2x3x3hy'=limh→02h+3x3+9x2h+9x h2+3h33x3hy'=limh→0h(9x2+9xh+3h2+2)h=9x2+23.y=ax+xf(x)=ax+xy'=limh→0f(x+h)f(x)hf(x+h)=ax+h+x+hy'=limh→0ax+h+x+haxxh+hh=limh→0ax+haxh+hh=limh→0ax(ah1)h+1limh→0(ah1)haxlimz→0z axln|1+z|lna=lnalimz→01ln|1+z|zaxah1=z¿lnalimz→01ln|1+z|1zax
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ah=1+z¿lnalimz→0axln|limz→0(1+z)1z|=lnalneaxhlna=ln|1+z|y'=axlna+1h=ln|1+z|lna4.
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