Question

Gostaria da Resolução da 5-C
Caso agilize, A derivada é: 1 / 2$\sqrt[5]{x} ^{4}$

Answer 1

$f\left(x \right )=\sqrt[5]{x}\;\Rightarrow\;f\left(x \right )=x^{\,^{1}\!\!\!\diagup\!\!_{5}}\\ \\ \\ f'\left(x \right )=\dfrac{1}{5}x^{\,^{1}\!\!\!\diagup\!\!_{5}-1}\\ \\ f'\left(x \right )=\dfrac{1}{5}x^{(1-5)/5}\\ \\ f'\left(x \right )=\dfrac{1}{5}x^{\,^{-4}\!\!\!\diagup\!\!_{5}}\\ \\ f'\left(x \right )=\dfrac{1}{5x^{\,^{4}\!\!\!\diagup\!\!_{5}}}\\ \\ f'\left(x \right )=\dfrac{1}{5\sqrt[5]{x^{4}}}$


Sendo assim, para $x=-32$, temos

$f'\left(-32 \right )=\dfrac{1}{5\sqrt[5]{\left(-32 \right )^{4}}}\\ \\ f'\left(-32 \right )=\dfrac{1}{5\sqrt[5]{\left(-1 \right )^{4}\cdot 32^{4}}}\\ \\ f'\left(-32 \right )=\dfrac{1}{5\sqrt[5]{1\cdot \left(2^{5} \right )^{4}}}\\ \\ f'\left(-32 \right )=\dfrac{1}{5\sqrt[5]{\left(2^{4} \right )^{5}}}\\ \\ f'\left(-32 \right )=\dfrac{1}{5\cdot 2^{4}}\\ \\ f'\left(-32 \right )=\dfrac{1}{5\cdot 16}\\ \\ f'\left(-32 \right )=\dfrac{1}{80}$