Question

Tính đạo hàm:

a) y=(x+1)tan4x

b) y=cos²(sin\(\sqrt{x}\))

c) y=tan(sin2x)

Answer 1

a/ \(y'=tan4x+\left(x+1\right)\left(tan4x\right)'=tan4x+\frac{4\left(x+1\right)}{cos^24x}\)

b/ \(y'=2cos\left(sin\sqrt{x}\right)\left[cos\left(sin\sqrt{x}\right)\right]'=2cos\left(sin\sqrt{x}\right).\left(-sin\left(sin\sqrt{x}\right)\right)\left(sin\sqrt{x}\right)'\)

\(=-\frac{sin\left(2sin\sqrt{x}\right).cos\sqrt{x}}{2\sqrt{x}}\)

c/ \(y'=\frac{\left(sin2x\right)'}{cos^2\left(sin2x\right)}=\frac{2cos2x}{cos^2\left(sin2x\right)}\)