Найдите производную 1. y = 5x^2 - 2x^3 + (3/5x) - 7 2. y = (2 корень из x) + 1\2 sinx -...

=========== 1 ===========
$y=5x^2 - 2x^3 + \frac{3}{5}x - 7\\ y'=(5x^2 - 2x^3 + \frac{3}{5}x - 7)'=(5x^2)' - (2x^3)' + (\frac{3}{5}x)' - (7)'=\\ =10x-6x^2+ \frac{3}{5}$
=========== 2 ===========
$y = 2 \sqrt{x} + \frac{1}{2} sinx - 3tgx\\ y'=(2 \sqrt{x} + \frac{1}{2} sinx - 3tgx)'=(2 \sqrt{x})' + (\frac{1}{2} sinx)' - (3tgx)'=\\ = \frac{1}{ \sqrt{x} } +\frac{1}{2}cosx- \frac{3}{cos^2x}$
=========== 3 ===========
$y= \sqrt{x} \cdot (5x-3)\\ y'=(\sqrt{x} \cdot (5x-3))'=(\sqrt{x})' \cdot (5x-3)+\sqrt{x} \cdot (5x-3)'=\\ = \frac{5x-3}{2 \sqrt{x} } +5 \sqrt{x}$
=========== 4 ===========
$y= \frac{x}{x^2+1} \\ y'=(\frac{x}{x^2+1})'=\frac{(x)'\cdot (x^2+1)-x\cdot(x^2+1)'}{(x^2+1)^2}= \frac{x^2+1-x\cdot2x}{(x^2+1)^2}= \frac{1-x^2}{(x^2+1)^2}$