Nếu \(f\left( x \right) = x\sin x\) thì \(f'\left( {\frac{{7\pi }}{2}} \right)\) bằng

* Hướng dẫn giải

\(f\left( x \right) = x\sin x\)

\( \Rightarrow f'\left( x \right) = \left( {x\sin x} \right)'\)\( = \left( x \right)'\sin x + x\left( {\sin x} \right)'\) \( = \sin x + x\cos x\)

Do đó \(f'\left( {\frac{{7\pi }}{2}} \right) = \sin \frac{{7\pi }}{2} + \frac{{7\pi }}{2}.\cos \frac{{7\pi }}{2}\)\( =  - 1 + \frac{{7\pi }}{2}.0 =  - 1\) .