Question

derivada de x³ + 3x / 3x+1

Answer 1

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: