Question

вычислить производную функции

Answer 1

$y '= \frac{(x - 2)'(2 {x}^{2} - 5x) - (2 {x}^{2} - 5x)' ( x- 2 )}{ {(2 {x}^{2} - 5x) }^{2} } = \\ = \frac{1 \times (2 {x}^{2} - 5x) - (4x - 5)(x - 2) }{ {(2 {x}^{2} - 5x)}^{2} } = \\ = \frac{2 {x}^{2} - 5x - 4 {x}^{2} + 8x + 5x - 10}{ {(2 {x}^{2} - 5x)}^{2} } = \\ = \frac{ - 2 {x}^{2} + 8x - 10}{ {(2 {x}^{2} - 5x)}^{2} } = - \frac{2 {x}^{2} - 8 + 10 }{ {(2 {x}^{2} - 5x)}^{2} }$

Answer 2

$\dfrac{-2x^{2}+8x-10}{(2x^{2}-5x)^{2}}$

Пошаговое объяснение:

$y=\dfrac{x-2}{2x^{2}-5x};$

$y'=\dfrac{(x-2)' \cdot (2x^{2}-5x)-(x-2) \cdot (2x^{2}-5x)'}{(2x^{2}-5x)^{2}}=\dfrac{2x^{2}-5x-(x-2) \cdot (4x-5)}{(2x^{2}-5x)^{2}}=$

$=\dfrac{2x^{2}-5x-(4x^{2}-5x-8x+10)}{(2x^{2}-5x)^{2}}=\dfrac{2x^{2}-4x^{2}-5x+13x-10}{(2x^{2}-5x)^{2}}=$

$=\dfrac{-2x^{2}+8x-10}{(2x^{2}-5x)^{2}};$