Đáp án:
Giải thích các bước giải:
\[\begin{align} & f(t)=\frac{\sin t-t\cos t}{\cos t-t\sin t} \\ & f'(t)={{\left( \frac{\sin t-t\cos t}{\cos t-t\sin t} \right)}^{'}} \\ & =\frac{(\sin t-t\cos t)'.(\cos t-t\sin t)-(\cos t-t\sin t)'.(\sin t-t\cos t)}{{{(\cos t-t\sin t)}^{2}}} \\ & =\frac{(c\text{os}t+t\sin t).(\cos t-t\sin t)+(\sin t+tc\text{os}t).(\sin t-t\cos t)}{{{(\cos t-t\sin t)}^{2}}} \\ & =\frac{{{\cos }^{2}}t-{{(t\sin t)}^{2}}+{{\sin }^{2}}t-{{(t\cos t)}^{2}}}{{{(\cos t-t\sin t)}^{2}}} \\ & =\frac{{{\cos }^{2}}t-{{t}^{2}}{{\sin }^{2}}t+{{\sin }^{2}}t-{{t}^{2}}{{\cos }^{2}}t}{{{(\cos t-t\sin t)}^{2}}} \\ & =\frac{{{\sin }^{2}}t+{{\cos }^{2}}t-{{t}^{2}}({{\sin }^{2}}t+{{\cos }^{2}}t)}{{{(\cos t-t\sin t)}^{2}}} \\ & =\frac{1-{{t}^{2}}}{{{(\cos t-t\sin t)}^{2}}} \\ & \Rightarrow f'(\pi )=\frac{1-{{\pi }^{2}}}{{{(\cos \pi -\pi \sin \pi )}^{2}}}=-32399 \\ \end{align}\]
