Question

[Vân Anh Phan]

Tính đạo hàm của HS

y= -cosx/3sin^3 + 4/3 cotx

Answer 1

[Bình Nhi]

y'=\(\dfrac{\left(-cosx\right)'.3.sin^3x-\left(-cosx\right)\left(3.sin^3x\right)'}{\left(3.sin^3x\right)^2}\)

y'=\(\dfrac{3.sinx.sin^3x+9.cosx.sin^2x}{^{ }\left(3.sin^3x\right)^2}\)

y'=\(\dfrac{3.sin^4x+9.cosx.sin^2x}{9.sin^6x}\)

y'=\(\dfrac{3.sin^2x\left(sin^2x+3.cosx\right)}{9.sin^6x}\)

y'=\(\dfrac{sin^2x+3.cosx}{3.sin^4x}\)

Answer 2

[Bình Nhi]

4/3cotx =\(\dfrac{-4}{3.sin^2x}\).