Question

Y'=((2x-1)e^x)'+3-(корень из х)'

Answer 1

Ответ:

$y = (2x - 1) {e}^{x} + 3 - \sqrt{x}$

$y' = (2x - 1) '{e}^{x} + ( {e}^{x} )'(2x - 1) + 0 - \frac{1}{2} {x}^{ - \frac{1}{2} } = \\ = 2 {e}^{x} + {e}^{x} \times (2x - 1) - \frac{1}{2 \sqrt{x} } = \\ = {e}^{ x} (2 + 2x - 1) - \frac{1}{2 \sqrt{x} } = {e}^{x} (2x + 1) - \frac{1}{2 \sqrt{x} }$