Question

8) Resolva está seguinte questão de Funções Derivadas !


$b) \frac{1}{3x^5} - 4\sqrt{x + 3} +5 \sqrt[5]{x^6-4 + \frac{5}{7x} }$

Answer 1

Note que a função digitada é levemente diferente da anexada. A resolução segue a original (anexo).

Vamos começar utilizando a propriedade da derivada da soma, ou seja, a derivada da soma pode ser calculada pela soma das derivadas de cada termo da função.

Sendo assim, podemos calcular cada termo separadamente para facilitar o calculo:

$\dfrac{d\left(\frac{1}{3x^5}\right)}{dx}~~\rightarrow~~Regra~do~Quociente\\\\\\\dfrac{d\left(\frac{1}{3x^5}\right)}{dx}~=~\dfrac{1'~.~(3x^5)~-~(3x^5)'~.~1}{\left(3x^5\right)^2}\\\\\\\dfrac{d\left(\frac{1}{3x^5}\right)}{dx}~=~\dfrac{0~.~(3x^5)~-~(5~.~3x^{4})~.~1}{9x^{10}}\\\\\\\dfrac{d\left(\frac{1}{3x^5}\right)}{dx}~=~\dfrac{-5~.~3x^{4}}{9x^{10}}\\\\\\\dfrac{d\left(\frac{1}{3x^5}\right)}{dx}~=~-\dfrac{3~.~5x^{4}}{9x^{10}}$

$\dfrac{d\left(\frac{1}{3x^5}\right)}{dx}~=~-\dfrac{5}{3x^{10-4}}\\\\\\\boxed{\dfrac{d\left(\frac{1}{3x^5}\right)}{dx}~=~-\dfrac{5}{3x^{6}}}$

$\dfrac{d\left(4\sqrt{x+3}~\right)}{dx}~=~4~.~\dfrac{d\left(\left(x+3\right)^{\frac{1}{2}}\right)}{dx}\\\\\\Utilizando~a~regra~da~cadeia\\\\\\u=x+3\\\\\\\dfrac{d\left(4\sqrt{x+3}~\right)}{dx}~=~4~.~\dfrac{d\left(u^{\frac{1}{2}}\right)}{du}~.~\dfrac{d(x+3)}{dx}\\\\\\\dfrac{d\left(4\sqrt{x+3}~\right)}{dx}~=~4~.~\left(\frac{1}{2}~.~u^{\frac{1}{2}-1}\right)~.~(1)\\\\\\\dfrac{d\left(4\sqrt{x+3}~\right)}{dx}~=~2~.~u^{-\frac{1}{2}}$

$\dfrac{d\left(4\sqrt{x+3}~\right)}{dx}~=~2~.~(x+3)^{-\frac{1}{2}}\\\\\\\boxed{\dfrac{d\left(4\sqrt{x+3}~\right)}{dx}~=~\dfrac{2}{\sqrt{x+3}}}$

$\dfrac{d\left(5\sqrt[5]{x^6-4x}~\right)}{dx}~=~5~.~\dfrac{d\left(\left(x^6-4x\right)^{\frac{1}{5}}\right)}{dx}\\\\\\Utilizando~a~regra~da~cadeia\\\\\\u=x^6-4x\\\\\\\dfrac{d\left(5\sqrt[5]{x^6-4x}~\right)}{dx}~=~5~.~\dfrac{d\left(u^{\frac{1}{5}}\right)}{du}~.~\dfrac{d\left(x^6-4x\right)}{dx}$

$\dfrac{d\left(5\sqrt[5]{x^6-4x}~\right)}{dx}~=~5~.~\frac{1}{5}u^{\frac{1}{5}-1}~.~\left(6x^5-4\right)\\\\\\\dfrac{d\left(5\sqrt[5]{x^6-4x}~\right)}{dx}~=~u^{-\frac{4}{5}}~.~\left(6x^5-4\right)\\\\\\\dfrac{d\left(5\sqrt[5]{x^6-4x}~\right)}{dx}~=~\left(x^6-4\right)^{-\frac{4}{5}}~.~\left(6x^5-4\right)$

$\dfrac{d\left(5\sqrt[5]{x^6-4x}~\right)}{dx}~=~\dfrac{1}{\sqrt[5]{\left(x^6-4x\right)^4}}~.~\left(6x^5-4\right)\\\\\\\boxed{\dfrac{d\left(5\sqrt[5]{x^6-4x}~\right)}{dx}~=~\dfrac{6x^5-4}{\sqrt[5]{\left(x^6-4x\right)^4}}}$

$\dfrac{d\left(\frac{5}{7x}\right)}{dx}~~~\rightarrow~~~Regra~do~Quociente$

$\dfrac{d\left(\frac{5}{7x}\right)}{dx}~=~\dfrac{5'~.~(7x)~-~(7x)'~.~5}{(7x)^2}\\\\\\\dfrac{d\left(\frac{5}{7x}\right)}{dx}~=~\dfrac{0~.~(7x)~-~7~.~5}{49x^2}\\\\\\\dfrac{d\left(\frac{5}{7x}\right)}{dx}~=~\dfrac{-35}{49x^2}\\\\\\\boxed{\dfrac{d\left(\frac{5}{7x}\right)}{dx}~=~-\dfrac{5}{7x^2}}$

Basta agora somar e/ou subtrair os termos respeitando os sinais da função.

$\dfrac{dn(x)}{dx}~=~-\left(-\dfrac{5}{3x^6}\right)~-~\left(\dfrac{2}{\sqrt{x+3}}\right)~+~\left(\dfrac{6x^5-4}{\sqrt[5]{\left(x^6-4x\right)^4}}}\right)~+~\left(-\dfrac{5}{7x^2}\right)\\\\\\\\\boxed{\dfrac{dn(x)}{dx}~=~\dfrac{5}{3x^6}~-~\dfrac{2}{\sqrt{x+3}}~+~\dfrac{6x^5-4}{\sqrt[5]{\left(x^6-4x\right)^4}}~-~\dfrac{5}{7x^2}}$