d) Xét hàm số y = sinxcosx

Cách 1.

y' = (sinxcosx)' = (sinx)'.cosx + sinx.(cosx)'

= cosx.cosx + sinx.(–sinx)

= cos²x – sin²x = cos2x.

Cách 2.

Ta có $y=sinxcosx=\frac{1}{2}\sin 2x.$

Suy ra $y^{\text{'}}={\left(\frac{1}{2}\sin 2x\right)}^{\text{'}}=\frac{1}{2}{\left(2x\right)}^{\text{'}}\cdot \cos 2x=\frac{1}{2}\cdot 2\cdot \cos 2x=\cos 2x.$

$y^{\text{'}}=\cos x\cdot \cos x-\sin x\cdot \sin x={\left(\cos x\right)}^2-{\left(\sin x\right)}^2=\cos 2x$