$f(t)=\dfrac{\sin t - t \cos t}{\cos t -t\sin t}\\ f'(t)=\dfrac{(\sin t - t \cos t)'(\cos t -t\sin t)-(\sin t - t \cos t)(\cos t -t\sin t)'}{\cos t -t\sin t}\\ =\dfrac{[\cos t - (t\cos t)'](\cos t -t\sin t)-(\sin t - t \cos t)[-\sin t -(t\sin t)']}{\cos t -t\sin t}\\ =\dfrac{[\cos t - (t'\cos t+t(\cos t)'))](\cos t -t\sin t)-(\sin t - t \cos t)[-\sin t -(t'\sin t+ t (\sin t)')]}{\cos t -t\sin t}\\ \dfrac{(\cos t - \cos t+t\sin t)(\cos t -t\sin t)-(\sin t - t \cos t)(-\sin t -\sin t -t\cos t)}{(\cos t -t\sin t)^2}\\ =\dfrac{t\sin t(\cos t -t\sin t)+(2\sin t+t\cos t)(\sin t - t \cos t)}{(\cos t -t\sin t)^2}\\ =\dfrac{t\sin t\cos t -t^2\sin^2t+2\sin^2t - 2t\sin t \cos t+t\cos t\sin t-t^2\cos^2t}{(\cos t -t\sin t)^2}\\ =\dfrac{-t^2+2\sin^2t}{(\cos t -t\sin t)^2}\\ f'(\pi)=\dfrac{-\pi^2+2\sin^2\pi}{(\cos \pi -\pi\sin \pi)^2}\\=-\pi^2$

Đáp án:

Giải thích các bước giải:

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