Question

[Julian Edward]

tìm y'

a) \(y=sin^3x\)

b) \(y=cos^3x\)

c) \(y=sinx.cos^2x\)

d) \(y=\sqrt[3]{x}+\sqrt[3]{\left(x+1\right)^2}\)

Answer 1

[Nguyễn Việt Lâm]

a. \(y'=3sin^2x.\left(sinx\right)'=3sin^2x.cosx\)

b. \(y'=3cos^2x.\left(cosx\right)'=-3cos^2x.sinx\)

c. \(y'=cosx.cos^2x+2cosx.\left(-sinx\right).sinx=cos^3x-2cosx.sin^2x\)

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