Question

y'=?
1)y=√(sinx-4)
2)y=tg8x

Answer 1

$1)\y=sqrt{sinx-4}\\y'=frac{1}{2sqrt{sinx-4}}cdot cosx=frac{cosx}{2sqrt{sinx-4}}\\2)\y=tg8x\\y'=frac{1}{cos8x}cdot8=frac{8}{cos8x}$