Question

Найти производную функции: $y = \frac{(8 {x}^{2} - 4x)^{2} }{5x}$

Answer 1

$y' = \frac{ {((8 {x}^{2} - 4x) }^{2})' \times 5x - {(8 {x}^{2} - 4x) }^{2} \times (5x )' }{ {(5x)}^{2} } =$

$= \frac{2(8 {x}^{2} - 4x)(16x - 4) \times 5x - (8 {x}^{2} - 4x) \times 5 }{25 {x}^{2} } =$

$= \frac{10x( 8{x}^{2} - 4x)(16x - 4) - 5(8 {x}^{2} - 4x)}{25 {x}^{2} } =$

$= \frac{1280 {x}^{4} - 960 {x}^{3} + 160 {x}^{3} - 5(8 {x}^{2} - 4x)^{2} }{25 {x}^{2} } =$

$= \frac{5(256 {x}^{4} - 192 {x}^{3} + 32 {x}^{2} - {(8 {x}^{2} - 4x)}^{2} }{25 {x}^{2} } =$

$= \frac{256 {x}^{4} - 192 {x}^{3} + 3 {x}^{2} - 64 {x}^{4} + 64 {x}^{3} - 16 {x}^{2} }{5 {x}^{2} } =$

$= \frac{192 {x}^{4} - 128 {x}^{3} + 16 {x}^{2} }{5 {x}^{2} } =$

$= \frac{ {x}^{2} (192 {x}^{2} - 128x + 16) }{5 {x}^{2} } = \frac{192 {x}^{2} - 128x + 16 }{5}$