Chapter 3 - Derivatives - 3.7 The Chain Rule - 3.7 Exercises - Page 194: 92

$${\text{The statements }}{\bf{a}}{\text{, }}{\bf{b}}{\text{, and }}{\bf{c}}{\text{ have been proved}}{\text{.}}$$

$$\eqalign{ & {\bf{a}}. \cr & \cos 2t = {\cos ^2}t - {\sin ^2}t \cr & {\text{Differentiate both sides with respect to }}t. \cr & \frac{d}{{dt}}\left[ {\cos 2t} \right] = \frac{d}{{dt}}\left[ {{{\cos }^2}t - {{\sin }^2}t} \right] \cr & - \sin 2t\left( 2 \right) = 2\cos t\left( { - \sin t} \right) - 2\sin t\left( {\cos t} \right) \cr & {\text{Simplifying}} \cr & - 2\sin 2t = - 2\sin t\cos t - 2\sin t\cos t \cr & - 2\sin 2t = - 2\left( {2\sin t\cos t} \right) \cr & {\text{Divide both sides by }} - 2 \cr & \frac{{ - 2\sin 2t}}{{ - 2}} = \frac{{ - 2\left( {2\sin t\cos t} \right)}}{{ - 2}} \cr & \sin 2t = 2\sin t\cos t \cr & \cr & {\bf{b}}. \cr & \cos 2t = 2{\cos ^2}t - 1 \cr & {\text{Differentiate both sides with respect to }}t. \cr & \frac{d}{{dt}}\left[ {\cos 2t} \right] = \frac{d}{{dt}}\left[ {2{{\cos }^2}t - 1} \right] \cr & - \sin 2t\left( 2 \right) = - 4\sin t\cos t - 0 \cr & {\text{Divide both sides by }} - 2 \cr & \frac{{ - \sin 2t\left( 2 \right)}}{{ - 2}} = \frac{{ - 4\sin t\cos t}}{{ - 2}} \cr & \sin 2t = 2\sin t\cos t \cr & \cr & {\bf{c}}. \cr & \sin 2t = 2\sin t\cos t \cr & {\text{Differentiate both sides with respect to }}t. \cr & \frac{d}{{dt}}\left[ {\sin 2t} \right] = \frac{d}{{dt}}\left[ {2\sin t\cos t} \right] \cr & 2\cos 2t = 2\sin t\left( { - \sin t} \right) + \left( {2\cos t} \right)\left( {\cos t} \right) \cr & {\text{Simplifying}} \cr & 2\cos 2t = - 2{\sin ^2}t + 2{\cos ^2}t \cr & 2\cos 2t = 2{\cos ^2}t - 2{\sin ^2}t \cr & {\text{Divide both sides by 2}} \cr & \cos 2t = {\cos ^2}t - {\sin ^2}t \cr} $$