Đáp án:
\[\dfrac{{{f_1}'\left( 1 \right)}}{{{f_2}'\left( 1 \right)}} = - 1\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{f_1}\left( x \right) = x.\sin x\\
\Rightarrow {f_1}'\left( x \right) = x'.\sin x + x.\left( {\sin x} \right)' = \sin x + x\cos x\\
\Rightarrow {f_1}'\left( 1 \right) = \sin 1 + 1.\cos 1 = \sin 1 + \cos 1\\
{f_2}\left( x \right) = \dfrac{{\cos x}}{x}\\
\Rightarrow {f_2}'\left( x \right) = \dfrac{{\left( {\cos x} \right)'.x - x'.\cos x}}{{{x^2}}} = \dfrac{{ - \sin x.x - \cos x}}{{{x^2}}}\\
\Rightarrow {f_2}'\left( 1 \right) = \dfrac{{ - \sin 1.1 - \cos 1}}{{{1^2}}} = - \left( {\sin 1 + \cos 1} \right)\\
\Rightarrow \dfrac{{{f_1}'\left( 1 \right)}}{{{f_2}'\left( 1 \right)}} = - 1
\end{array}\)
