Question

Tính đạo hàm : y=✓sin⁴x+cos⁴x

Answer 1

\(y'=\dfrac{\left(sin^4x+cos^4x\right)'}{2\sqrt{sin^4x+cos^4x}}=\dfrac{2sin^3x.cosx-2cos^3x.sinx}{\sqrt{sin^4x+cos^4x}}\)