Question

Найти для функции y'x и y''x

Answer 1

Ответ:

$y = {x}^{4} ln(4x)$

$y' = ( {x}^{4} )' ln(4x) + ( ln(4x))' \times (4x)' \times {x}^{4} = \\ = 4 {x}^{3} ln(4x) + \frac{1}{4x} \times 4 \times {x}^{4} = \\ = 4 {x}^{3} ln(4x) + {x}^{3} = \\ = {x}^{3} (4 ln(4x) + 1)$

$y'' = ( {x}^{3} )' \times (4 ln(4x) + 1) + (4 ln(4x) + 1) '\times ( ln(4x)) ' \times (4x)' \times {x}^{3} = \\ = 3 {x}^{2} (4 ln(4x) + 1) + \frac{4}{4x} \times 4 \times {x}^{3} = \\ = 3 {x}^{2} (4 ln(4x) + 1) + 4 {x}^{2} = \\ = {x}^{2} (12 ln(4x) + 3 + 4) = \\ = {x}^{2} (12 ln(4x) + 7)$