2. If $p=\frac{q^2}{\pi },q=3\cos (r)+\sin (r),r=\ln \left(s^{\pi }\right)$ find $\frac{dp}{ds}$ at $s=e$. (Answer must be exact. 6 marks)

Step 1 ：First, substitute the expression for r into the expression for q: $q=3\cos (\ln (s^{\pi }))+\sin (\ln (s^{\pi }))$

Step 2 ：Now, substitute the expression for q into the expression for p: $p=\frac{(3\cos (\ln (s^{\pi }))+\sin (\ln (s^{\pi })){)}^2}{\pi }$

Step 3 ：To find $\frac{dp}{ds}$, we need to differentiate p with respect to s: $\frac{dp}{ds}=\frac{d}{ds}\left(\frac{(3\cos (\ln (s^{\pi }))+\sin (\ln (s^{\pi })){)}^2}{\pi }\right)$

Step 4 ：Apply the chain rule: $\frac{dp}{ds}=\frac{2(3\cos (\ln (s^{\pi }))+\sin (\ln (s^{\pi })))}{\pi }\cdot \frac{d}{ds}(3\cos (\ln (s^{\pi }))+\sin (\ln (s^{\pi })))$

Step 5 ：Differentiate the expression inside the parentheses with respect to s: $\frac{d}{ds}(3\cos (\ln (s^{\pi }))+\sin (\ln (s^{\pi })))=-3\sin (\ln (s^{\pi }))\cdot \frac{\pi }{s}+\cos (\ln (s^{\pi }))\cdot \frac{\pi }{s}$

Step 6 ：Substitute this expression back into the expression for $\frac{dp}{ds}$: $\frac{dp}{ds}=\frac{2(3\cos (\ln (s^{\pi }))+\sin (\ln (s^{\pi })))}{\pi }\cdot (-3\sin (\ln (s^{\pi }))\cdot \frac{\pi }{s}+\cos (\ln (s^{\pi }))\cdot \frac{\pi }{s})$

Step 7 ：Now, substitute s = e into the expression for $\frac{dp}{ds}$: $\frac{dp}{ds}=\frac{2(3\cos (\ln (e^{\pi }))+\sin (\ln (e^{\pi })))}{\pi }\cdot (-3\sin (\ln (e^{\pi }))\cdot \frac{\pi }{e}+\cos (\ln (e^{\pi }))\cdot \frac{\pi }{e})$

Step 8 ：Simplify the expression: $\frac{dp}{ds}=\frac{2(3\cos (\pi )+\sin (\pi ))}{\pi }\cdot (-3\sin (\pi )\cdot \frac{\pi }{e}+\cos (\pi )\cdot \frac{\pi }{e})$

Step 9 ：Evaluate the trigonometric functions: $\frac{dp}{ds}=\frac{2(-3+0)}{\pi }\cdot (0-\frac{\pi }{e})$

Step 10 ：Finally, simplify the expression: $\frac{dp}{ds}=\boxed{{\displaystyle \frac{6\pi }{e}}}$