Bài 1 trang 48

Đề bài

Tính đạo hàm của các hàm số sau:

a) $y=2x^3-\frac{x^2}{2}+4x-\frac{1}{3}$;

b) $y=\frac{-2x+3}{x-4}$;

c) $y=\frac{x^2-2x+3}{x-1}$;

d) $y=\sqrt{5x}$.

Lời giải chi tiết

a) $y^{\prime }=2.3x^2-\frac{1}{2}.2x+4.1-0=6x^2-x+4$.

b) $y^{\prime }=\frac{{\left(-2x+3\right)}^{\prime }.\left(x-4\right)-\left(-2x+3\right).{\left(x-4\right)}^{\prime }}{{\left(x-4\right)}^2}$

$=\frac{-2\left(x-4\right)-\left(-2x+3\right).1}{{\left(x-4\right)}^2}$

$=\frac{-2x+8+2x-3}{{\left(x-4\right)}^2}=\frac{5}{{\left(x-4\right)}^2}$.

c) $y^{\prime }=\frac{{\left(x^2-2x+3\right)}^{\prime }\left(x-1\right)-\left(x^2-2x+3\right){\left(x-1\right)}^{\prime }}{{\left(x-1\right)}^2}$

$=\frac{\left(2x-2\right)\left(x-1\right)-\left(x^2-2x+3\right).1}{{\left(x-1\right)}^2}$ $=\frac{2x^2-2x-2x+2-x^2+2x-3}{{\left(x-1\right)}^2}$

$=\frac{x^2-2x-1}{{\left(x-1\right)}^2}$.

d) $y^{\prime }={\left(\sqrt{5}.\sqrt{x}\right)}^{\prime }=\sqrt{5}.\frac{1}{2\sqrt{x}}=\frac{\sqrt{5}}{2\sqrt{x}}=\frac{5}{2\sqrt{5x}}$.