Question

Найдите значение производной функции в указанной точке.
f(x) =$sqrt[5]{x^4}$ , x0 = 32.

Answer 1

$it f(x) = sqrt[5]{x^4} = x^{frac{4}{5}} \;\ f'(x) = dfrac{4}{5} x^{frac{4}{5}-1} = dfrac{4}{5} x^{-frac{1}{5}} =dfrac{4}{5}cdot dfrac{1}{x^frac{1}{5}} =dfrac{4}{5x^frac{1}{5}} \;\ \;\ f'(32) =dfrac{4}{5cdot 32^frac{1}{5}} = dfrac{4}{5cdot(2^5)^frac{1}{5}}=dfrac{4}{5cdot2} = dfrac{2}{5} =0,4$