(a) If y = cos^3( sec^2 2 t ), find dy/dt.

$$\begin{aligned} & \mathrm{y}=\cos ^{3}\left(\sec ^{2} 2 \mathrm{t}\right) \\ & \Rightarrow \frac{d \mathrm{y}}{d \mathrm{t}}=3 \cos ^{2}\left(\sec ^{2} 2 \mathrm{t}\right)\left[-\sin \left(\sec ^{2} 2 \mathrm{t}\right)\right] \times \frac{d\left(\sec ^{2} 2 \mathrm{t}\right)}{d \mathrm{t}} \\ & \Rightarrow \frac{d \mathrm{y}}{d \mathrm{t}}=-3 \cos ^{2}\left(\sec ^{2} 2 \mathrm{t}\right) \cdot \sin \left(\sec ^{2} 2 \mathrm{t}\right) \times 2 \sec 2 \mathrm{t} \cdot \sec 2 \mathrm{t} \tan 2 \mathrm{t} \cdot 2 \\ & \therefore \frac{d \mathrm{y}}{d \mathrm{t}}=-12 \cos ^{2}\left(\sec ^{2} 2 \mathrm{t}\right) \times \sin \left(\sec ^{2} 2 \mathrm{t}\right) \times \sec ^{2} 2 \mathrm{t} \times \tan 2 \mathrm{t} . \end{aligned} $$ $$\begin{aligned} & A s, x^{\mathrm{y}}=\mathrm{e}^{x-\mathrm{y}} \Rightarrow \log \left(x^{\mathrm{y}}\right)=\log \left(\mathrm{e}^{x-\mathrm{y}}\right) \\ & \Rightarrow \mathrm{y} \log x=(x-\mathrm{y}) \Rightarrow \mathrm{y}=\frac{x}{1+\log x} \end{aligned} $$ Now, Differentiating both the sides wrt $\displaystyle x$ $$\frac{d \mathrm{y}}{d x}=\frac{(\log x+1) \cdot 1-x\left(\dfrac{1}{x}\right)}{(\log x+1)^{2}}=\frac{\log x}{(1+\log x)^{2}} $$