Question

Tính đạo hàm : y=sin⁴xcosx

Answer 1

\(y'=\left(\sin^4x\right)'\cos x+\sin^4x\left(\cos x\right)'=4\sin^3x.\cos^2x-\sin^5x\)

Answer 2

\(y'=4sin^3x.cosx.cosx-sinx.sin^4x=4sin^3x.cos^2x-sin^5x\)