# Atps de calculo 2 periodo

Páginas: 4 (802 palavras) Publicado: 14 de abril de 2013
TAtps de calculo
1) f’(x) sendo:
a) f(x) =7x+3
f'x=∆x→0 ∆x lim f x+∆x- f (x)
lim 7x+∆x+3-(7x+3) ∆x→0 ∆xlim7x+7∆x+3-7x-3∆x→0∆x
xlim7∆x∆x→0∆x
∆→0lim 7
f'x=7
b) f(x) =3x²+4 f'x=∆x→0 ∆xlim f x+∆x- f (x)
lim3x+∆x2+4-3x²+4∆x→0∆x
lim3x2+2.x.∆x+∆x2+4-3x²+4∆x→0∆x
lim3x²+6x.∆x+3x²+4-3x²-4∆x→0∆x
∆x→0 lim⁡ 6x
f'x=6x
c) f(x) =3x²-2x+1
f'x=∆x→0∆x lim f x+∆x- f (x) lim3x+∆x2-2x+∆x+1-(3x2-2x+1)∆x→0∆x lim3x2+2x.∆x+∆x2-2x-2∆x+1-(3x2-2x+1)∆x→0∆x lim3x²+6x∆x+3∆x²-2x-2∆x+1-3x²+2x-1∆x→0∆xlim6x∆x+3∆x²-2∆x∆x→0∆x
lim∆x(6x∆x+3∆x2-2∆x)∆x→0∆x
lim6x+3∆x-2∆x→0
∆x→0lim 6x-2 f'x=6x-2

2) Nos exercícios abaixo calcule a derivada aplicando os teoremas.
a) f(x) = (x²-3x+2).(2x³+1)h’(x)= f(x).g’(x)+g(x).f’(x)
f’(x)= (x²-3x+2).(6x²)+(2x³+1).(2x-3)
f’x= 6x4-18x³+12x²+4x4-6x³+2x-3
f’(x)= 10x4-24³+12x²+2x-3 |

b) f(x)= x³-8x³+8
h'x=gx.f'x-fx.g'(x)[gx]²f'x=x3+8.3x2-x3-8.(3x²)(x3+8)² f'x=x3+8.3x2-x3-8.(3x2)x6+16x³+64
f'x=3x5+24x²-3x5+24x²x6+16x³+64
f'x=48x²x6+16x³+64 |

c) fx=x²-2x+1x²+2x+1
f'x=gx.f'x-fx.g'(x)[gx]²f'x=x2-2x+1.2x+2-x2+2x+1.(2x-2)(x2-2x+1)² f'x=x2-2x+1.2x+2-x2+2x+1.(2x-2)x2-2x+1.(x2-2x+1) f'x=2x³+2x²-4x²-4x+2x+2-2x³+2x²-4x²+4x-2x+1x4-2x³+x²-2x³+4x²-2x+x²-2x+1
f'x=-4x²+3x4-4x³+6x²-4x+1 |

d) fx=x4x4-2x²+5x+1
h'x=gx.f'x-fx.g'(x)[gx]² f'x=x4.4x3-4x+5-x4-2x2+5x+1.(4x³)(x4)² f'x=4x7-4x5+5x4-4x7+8x5-20x4-4x³x8
f'x=4x5-15x4-4x³ x8 |

e) fx=x3-2x+1.(2x2+3x)h’(x)= f(x).g’(x)+g(x).f’(x)
f’ (x)= (4x3-2x+1).(4x+3)+(2x²+3x).(3x²-2)
f’(x)= 4x4+3x³-8x²-6x+4x+3+6x4-4x²+9x³-6x
f’(x)= 10x4+12x³-12x²-8x+3 |

f) fx=(4x2+3)4
fx=4x2+32.4x2+32...

Por favor, assinar para o acesso.