Question

Найти производную y=√x(-2x+1)

Answer 1

$y'=\sqrt{x}\left(-2x+1\right)\:\\\\\frac{d}{dx}\left(\sqrt{x}\left(-2x+1\right)\right) = \\=\frac{d}{dx}\left(\sqrt{x}\right)\left(-2x+1\right)+\frac{d}{dx}\left(-2x+1\right)\sqrt{x} =\\=\frac{d}{dx}\left(x^{\frac{1}{2}}\right)\left(-2x+1\right)+ (-\frac{d}{dx}\left(2x\right)+\frac{d}{dx}\left(1\right))\sqrt{x} =\\\frac{1}{2}x^{\frac{1}{2}-1}\left(-2x+1\right)+(-2+0)\sqrt{x}=\\| \:\: \frac{1}{2}x^{\frac{1}{2}-1} =\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2}\cdot \frac{1}{\sqrt{x}}=\frac{1}{2\sqrt{x}}$

$=\frac{1}{2\sqrt{x}}\left(-2x+1\right)+\left(-2\right)\sqrt{x} = \\=\frac{-2x+1}{2\sqrt{x}}-2\sqrt{x}=\\=\frac{-2x+1-2\sqrt{x}\cdot \:2\sqrt{x}}{2\sqrt{x}}=\\=\frac{-6x+1}{2\sqrt{x}}$