Найдите производную функций номер 748(в,г) 750(а,б)

748
в)
$y=\sqrt{x}(8x-10);\\ y'=\left(\sqrt{x}\right)'\cdot\left(8x-10\right)+\sqrt{x}\cdot\left(8x-10\right)'=\\ =\frac{1}{2\sqrt{x}}(8x-10)+\sqrt{x}\cdot8=4\sqrt{x}-\frac{5}{\sqrt{x}}+8\sqrt{x}=\\ =12\sqrt{x}-\frac{5}{\sqrt{x}}=\frac{12x-5}{\sqrt x}$
г)
$y=\sqrt{x}(x^4+2);\\ y'=\left(\sqrt{x}\right)'\cdot\left(x^4+2\right)+\sqrt{x}\cdot\left(x^4+2\right)'=\\ =\frac{1}{2\sqrt{x}}(x^4+2)+\sqrt{x}\cdot4\cdot x^3=\frac{x^3\sqrt{x}}{2}+\frac{1}{\sqrt{x}}+4x^3\sqrt{x}=\\ =4\frac12x^3\sqrt{x}+\frac{1}{\sqrt{x}}=\frac{9x^4+2}{2\sqrt{x}}$

750

а)

$y=\left(\frac1x+1\right)\left(2x-3\right);\\ y'=\left(\frac1x+1\right)'\cdot\left(2x-3\right)+\left(\frac1x+1\right)\cdot\left(2x-3\right)'=\\ =-\frac1{x^2}\cdot\left(2x-3\right)+\left(\frac1x+1\right)\cdot2=\\ =-\frac2x+\frac3{x^2}+\frac2x+2=2+\frac{3}{x^2}$

б)

$y=\left(6-\frac1x\right)\left(6x+1\right);\\ y'=\left(6-\frac1x\right)'\cdot\left(6x+1\right)+\left(6-\frac1x\right)\cdot\left(6x+1\right)'=\\ =\frac1{x^2}\cdot\left(6x+1\right)+\left(6-\frac1x\right)\cdot6=\\ =\frac6x+\frac1{x^2}+36-\frac6x=\frac{1}{x^2}+36$