Với mọi x sao cho \(\cos x - x\sin x \ne 0,\) ta có:
\(f'\left( x \right) = {{\left[ {\cos x - \left( {\cos x - x\sin x} \right)} \right]\left( {\cos x - x\sin x} \right) - \left( {\sin x - x\cos x} \right)\left[ { - \sin x - \left( {\sin x + x\cos x} \right)} \right]} \over {{{\left( {{\mathop{\rm cosx}\nolimits} - xsinx} \right)}^2}}}\)
Vì \(\sin \pi = 0,\cos \pi = - 1\) nên : \(f'\left( \pi \right) = {{\left[ { - 1 - \left( { - 1} \right)} \right].\left( { - 1} \right) - \pi .\pi } \over {{{\left( { - 1} \right)}^2}}} = - {\pi ^2}\)
