# Câu 42 trang 218 SGK Đại số và Giải tích 11 Nâng cao

##### Hướng dẫn giải

a. Ta có:

$$\begin{array}{l} f'\left( x \right) = 4{x^3} + 2\sin 2x\\ f"\left( x \right) = 12{x^2} + 4\cos 2x\\ {f^{\left( 3 \right)}} = 24x - 8\sin 2x\\ {f^{\left( 4 \right)}}\left( x \right) = 24 - 16\cos 2x \end{array}$$

b.

$$\begin{array}{l} f'\left( x \right) = 2\cos x\left( { - \sin x} \right) = - \sin 2x\\ f"\left( x \right) = - 2\cos 2x\\ {f^{\left( 3 \right)}}\left( x \right) = 4\sin 2x\\ {f^{\left( 4 \right)}} = 8\cos 2x\\ {f^{\left( 5 \right)}}\left( x \right) = - 16\sin 2x \end{array}$$

c.

$$\begin{array}{l} f'\left( x \right) = 6{\left( {x + 10} \right)^5}\\ f"\left( x \right) = 30{\left( {x + 10} \right)^4}\\ {f^{\left( 3 \right)}}\left( x \right) = 120{\left( {x + 10} \right)^3}\\ {f^{\left( 4 \right)}}\left( x \right) = 360{\left( {x + 10} \right)^2}\\ {f^{\left( 5 \right)}}\left( x \right) = 720\left( {x + 10} \right)\\ {f^{\left( 6 \right)}}\left( x \right) = 720\\ {f^{\left( n \right)}}\left( x \right) = 0,\forall n \ge 7 \end{array}$$