Cho \(f_1(x)=\frac{cosx}{x}, f_2(x)=xsinx.\) Tính \(\frac{f_1(1)}{f_2(1)}.\)
Ta có \({f_1}'(x) = \left( {\frac{{\cos x}}{x}} \right)' = \frac{{{{(cosx)}^\prime }.x - cosx.x'}}{{{x^2}}} = \frac{{ - x.sinx - cosx}}{{{x^2}}}.\)
\(f'_2(x)=(xsinx)'=x^2.sinx+x(sinx)'=sinx+xcosx.\)
Suy ra \(f'_1(1)=-sin1-cos1.\)
\(f'_2(1)=sin1+cos1.\)
\(\frac{f'_1(1)}{f'_2(1)}=\frac{-sin1-cos1}{sin1+cos1}=-1.\)
-- Mod Toán 11