Cách làm cho bạn:
\({f_1}\left( x \right) = {{\cos x} \over x};{f_2}\left( x \right) = x\sin x\)
Ta có:
\({f_1}'(x) = {{ - x.\sin x - \cos x} \over {{x^2}}} \Rightarrow {f_1}'(1) = - \sin 1 - \cos 1 = - (\sin 1 + \cos 1) \)
\({f_2}'(x) = \sin x + x.cosx \Rightarrow {f_2}'(1) = \sin 1 + \cos 1 \)
\(\Rightarrow {{{f_1}'(1)} \over {{f_2}'(1)}} =\frac{- (\sin 1 + \cos 1)}{\sin 1 + \cos 1} =- 1 \)
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