Question

[Nguyễn Phương Thùy]

tính đạo hàm của các hàm số sau

a, y=\(-\dfrac{3x^4}{8}+\dfrac{2x^3}{5}-\dfrac{x^2}{2}+5x-2021\)

b, y= \(\sqrt{x^2+4x+5}\)

c, y=\(\sqrt[3]{3x-2}\)

d, y=(2x-1)\(\sqrt{x+2}\)

e, y=\(sin^3\left(\dfrac{\pi}{3}-5x\right)\)

g, y=\(cot^{^4}\left(\dfrac{\pi}{6}-3x\right)\)

Answer 1

[Nguyễn Việt Lâm]

a.

\(y'=-\dfrac{3}{2}x^3+\dfrac{6}{5}x^2-x+5\)

b.

\(y'=\dfrac{\left(x^2+4x+5\right)'}{2\sqrt{x^2+4x+5}}=\dfrac{2x+4}{2\sqrt{x^2+4x+5}}=\dfrac{x+2}{\sqrt{x^2+4x+5}}\)

c.

\(y=\left(3x-2\right)^{\dfrac{1}{3}}\Rightarrow y'=\dfrac{1}{3}\left(3x-2\right)^{-\dfrac{2}{3}}=\dfrac{1}{3\sqrt[3]{\left(3x-2\right)^2}}\)

d.

\(y'=2\sqrt{x+2}+\dfrac{2x-1}{2\sqrt{x+2}}=\dfrac{6x+7}{2\sqrt{x+2}}\)

e.

\(y'=3sin^2\left(\dfrac{\pi}{3}-5x\right).\left[sin\left(\dfrac{\pi}{3}-5x\right)\right]'=-15sin^2\left(\dfrac{\pi}{3}-5x\right).cos\left(\dfrac{\pi}{3}-5x\right)\)

g.

\(y'=4cot^3\left(\dfrac{\pi}{6}-3x\right)\left[cot\left(\dfrac{\pi}{3}-3x\right)\right]'=12cot^3\left(\dfrac{\pi}{6}-3x\right).\dfrac{1}{sin^2\left(\dfrac{\pi}{3}-3x\right)}\)