Matemática, perguntado por ednadossantos, 5 meses atrás

derivada de x³ + 3x / 3x+1

Soluções para a tarefa

Respondido por Kin07
0

Resposta:

Solução:

\displaystyle \sf  \dfrac{d}{dx} \: \bigg[\dfrac{x^3+3x}{3x + 1} \bigg]

Regra do quociente:

\boxed{\displaystyle \sf  \dfrac{d}{dx} \: \bigg[\dfrac{f(x)}{g(x)} \bigg] =  \dfrac{\dfrac{d}{dx} \:[f(x)] \cdot g(x) - f(x) \cdot \dfrac{d}{dx} \:[g(x)]}{\big[g(x)\big]^2}  }

Resolvendo:

\displaystyle \sf  f(x) = x^{3} +3x \Rightarrow f'(x)  = 3x^{2} +3

\displaystyle \sf  g(x) = 3x+ 1 \Rightarrow g'(x)  = 3

\displaystyle \sf  \dfrac{d}{dx} \: \bigg[\dfrac{f(x)}{g(x)} \bigg] =  \dfrac{\dfrac{d}{dx} \:[f(x)] \cdot g(x) - f(x) \cdot \dfrac{d}{dx} \:[g(x)]}{\big[g(x)\big]^2}

\displaystyle \sf  \dfrac{d}{dx} \: \bigg[\dfrac{x^{3}+3x}{3x+1} \bigg] =  \dfrac{(3x^{2} +3) \cdot (3x+1) - (x^3+3x) \cdot 3}{ (3x+ 1)^2}

\displaystyle \sf  \dfrac{d}{dx} \: \bigg[\dfrac{x^{3}+3x}{3x+1} \bigg] =  \dfrac{9x^3+3x^{2} +9x +3 - (3x^3+9x) }{ (3x + 1)^2}

\displaystyle \sf  \dfrac{d}{dx} \: \bigg[\dfrac{x^{3}+3x}{3x+1} \bigg] =  \dfrac{9x^3+3x^{2} +\diagup\!\!\!{   9x} +3 - 3x^3- \diagup\!\!\!{  9x }}{ (3x + 1)^2}

\displaystyle \sf  \dfrac{d}{dx} \: \bigg[\dfrac{x^{3}+3x}{3x+1} \bigg] =  \dfrac{9x^3 -  3x^3+3x^{2} +3 }{ (3x + 1)^2}

\boxed{ \boxed { \boldsymbol{ \displaystyle \sf  \dfrac{d}{dx} \: \bigg[\dfrac{x^{3}+3x}{3x+1} \bigg] =  \dfrac{6x^3 +3x^{2} +3 }{ (3x +  1)^2}  }}} \quad \gets \text{\sf \textbf{Resposta  } }

Explicação passo a passo:

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