Question

100 PONTOS A derivada de $f(x)=\frac{(2x^{3}+1)^{32}}{x+2}$

Answer 1

Resposta:

Método do quociente

f =u / v

f' = (u'*v -u*v')/v²

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f(x)=(2x³+1)³²/(x+2)

f'(x)={ [(2x³+1)³²]' *(x+2) - (2x³+1)³² * (x+2)' }/(x+2)²

f'(x)={ [32*(2x³+1)³¹*6x²] *(x+2) - (2x³+1)³² * (1) }/(x+2)²

f'(x)={ [192*x²*(2x³+1)³¹] *(x+2) - (2x³+1)³² }/(x+2)²

f'(x)={(2x³+1)³¹*(192x³+384x²-2x³-1 }/(x+2)²

f'(x)= (2x³+1)³¹*(190x³+384x²-1 )/(x+2)²

Answer 2

• Oie tudo bom☺?!

• Resolvendo pelo Método do Quociente temos que:

$f= \frac{u}{v}$

$f'= (u'.v-u.v') / v^2$

• Agora resolvendo a questão:

$f(x)=(2x^3+1)^3^2/(x+2)\\\\f'(x)={ [(2x^3+1)^3^2]' .(x+2) - (2x^3+1)^3^2 . (x+2)' }/(x+2)^2\\\\f'(x)={ [32.(2x^3+1)^3^1.6x^2] .(x+2) - (2x^3+1)^3^2 . (1) }/(x+2)^2\\\\f'(x)={ [192.x^2.(2x^3+1)^3^1] .(x+2) - (2x^3+1)^3^2 }/(x+2)^2\\\\f'(x)={(2x^3+1)^3^1.(192x^3+384x^2-2x^3-1 }/(x+2)^2\\\\\boxed{\boxed{f'(x)= (2x^3+1)^3^1.(190x^3+384x^2-1 )/(x+2)^2}}$

Att↔ Trasherx❄☥✎Ⓐ