Question

Найдите производную функции $f(x)=\sqrt[3]{2t-t^{2} }$ и вычислите $f'(4)$

Answer 1

Ответ:

$f(t) = \sqrt[3]{2 t- {t}^{2} }$

$f'(t) = \frac{1}{3} {(2t - {t}^{2}) }^{ - \frac{2}{3} } \times (2t - t {}^{2} ) '= \\ = \frac{2 - 2t}{3 \sqrt[3]{2t - {t}^{2} } } = \frac{2(1 -t )}{3 \sqrt[3]{2t - {t}^{2} } }$

$f(4) = \frac{2(1 - 4)}{3 \sqrt[3]{2 \times 4 - {4}^{2} } } = \frac{2 \times ( - 3)}{3 \sqrt[3]{8 - 16} } = \\ = - \frac{2}{ \sqrt[3]{ - 8} } = - \frac{2}{( - 2)} = 1$