(a) If x = e^cos 3 t and y = e^sin 3 t, prove that dy/dx=-(y log x)/(x log y).

$$\begin{aligned} & \frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \mathrm{t}}=\mathrm{e}^{\cos 3 \mathrm{t}} \times(-\sin 3 \mathrm{t}) \times 3 \\ & \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} \mathrm{t}}=\mathrm{e}^{\sin 3 \mathrm{t}} \times(\cos 3 \mathrm{t}) \times 3 \end{aligned} $$ $$\begin{aligned} & \mathrm{x}=\mathrm{e}^{\cos 3 \mathrm{t}} \Rightarrow \cos 3 \mathrm{t}=\log \mathrm{x} \\ & \mathrm{y}=\mathrm{e}^{\sin 3 \mathrm{t}} \Rightarrow \sin 3 \mathrm{t}=\log \mathrm{y} \\ & \therefore \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} \mathrm{x}}=\frac{-\mathrm{y} \log \mathrm{x}}{\mathrm{x} \log \mathrm{y}} \end{aligned} $$ $$\begin{aligned} & \frac{\mathrm{d}(|\mathrm{x}|)}{\mathrm{d} \mathrm{x}}=\frac{\mathrm{d}\left(\sqrt{\mathrm{x}^{2}}\right)}{\mathrm{d} \mathrm{x}}, \mathrm{x} \neq 0 \\ & =\frac{1}{2}\left(\mathrm{x}^{2}\right)^{-\frac{1}{2}} \times \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{d} \mathrm{x}} \\ & =\frac{1}{2 \sqrt{\mathrm{x}^{2}}} 2 \mathrm{x}=\frac{\mathrm{x}}{|\mathrm{x}|} \end{aligned} $$