Đáp án đúng: A
Điều kiện: $ \sin 4x
e 0 $ $ f(x)=\dfrac{{{\cot }^ 2 }2x-1}{2\cot 2x}=\dfrac{1}{2} \left( \cot 2x-\dfrac{1}{{}\cot 2x} \right)=\dfrac{1}{2} \left( \dfrac{\cos 2x}{\sin 2x}-\dfrac{\sin 2x}{\cos 2x} \right)=\dfrac{\cos 4x}{\sin 4x}=\cot 4x. $ $ \begin{align} & f '(x)=-4(1+{{\cot }^ 2 }4x)=-4-4{{\cot }^ 2 }4x\,; \\ & f ''(x)=32\cot 4x(1+{{\cot }^ 2 }4x)=32cot4x+32co{ t ^ 3 }4x\,; \\ & f '''(x)=-128(1+{{\cot }^ 2 }4x)+384{{\cot }^ 2 }4x.(-1-{{\cot }^ 2 }4x) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-128+256{{\cot }^ 2 }4x+384{{\cot }^ 4 }4x \\ \end{align} $ $ \begin{align} & f '''(x)+64 f '(x)=-128+256{{\cot }^ 2 }4x+384{{\cot }^ 4 }x+64\left( -4-4{{\cot }^ 2 }4x \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=384\left( {{\cot }^ 4 }x-1 \right)=384\left( { f ^ 2 }(x)-1 \right) \\ \end{align} $